Sketch the graph of a function
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Hi,
For problem 5, y²(x² - 9) = 2x, we can note the following:
Domain:
Unless otherwise mentioned, in elementary mathematics, y would be the range variable and x would be the domain variable.
The domain of this equation would thus be the set of all x values in the solution set of the equation (the solution set of this equation is the set of all ordered pairs, (x, y), that solve the equation).
We usually find this set by examining the operations involved in the equation, since most operations can be done with almost any number, except for a limited known set of numbers. Thus, our thinking is exclusionary in nature, not inclusionary.
In particular, both the left and right sides of this equation are polynomials. Polynomials only involve addition and multiplication, which is defined for all numbers, and thus do not exclude any values of x.
However, since this equation is not solved in a manner that allows us to easily see how y is related to x, we should then ask ourselves if there is any value of x that is not associated with any value of y (such a value of x would not be in the solution set of the equation).
In particular, notice that if we try to solve this equation using algebra for y as a function of x, we would have to divide both sides of the equation by the value of (x² - 9).
Division is an operation that is only defined for non-zero numbers. So if (x² - 9) is non-zero, we have no problems and we can solve for y as a function of whatever x is. However, we should inspect what happens if (x² - 9) is 0.
This occurs when x is either 3 or -3 (solve the equation x² - 9 = 0 for x). When x is either 3 or -3, the original equation becomes either 0 = 6 or 0 = -6, both of which are false statements. Thus, the values of 3 and -3 are not in the domain of this equation, since they are not associated with any solution of the equation.
Let us keep trying to solve for y as a function of x.
If x is neither 3 nor -3, then \(y^2 = \frac{2x}{x² - 9}\), and the Square Root Property then tells us that the only possible values of y are \(y = \pm\sqrt{\frac{2x}{x^2 - 9}}\).
Notice the principal square root function, √. In the Cartesian plane, where these equations are usually graphed, we only have real numbers. When the number inside the √ function is negative, the result is no longer a real number. When the number is nonnegative, we always get a real result.
Thus, if we are interested in domains of graphs in the Cartesian plane, we have to throw out any values of x that make the value of \(\frac{2x}{x^2 - 9}\) negative.
There are two ways that a fraction can be negative: the numerator is positive and the denominator is negative, or the denominator is negative and the numerator is positive.
Let us examine the first system of inequalities: 2x > 0 and x² - 9 < 0.
This system is solved when x > 0 and -3 < x < 3, which can be simplified to the single condition 0 < x < 3.
Thus, we know that all values of x in the interval where 0 < x < 3 are not in the domain of the equation.
Let us examine the second system of inequalities: 2x < 0 and x² - 9 > 0.
By similar algebra, we find the solution set of this system of inequalities is the interval x < -3.
Thus, having isolated y in terms of x, we see that when x = 3, x = -3, 0 < x < 3, or x < -3, the original equation has no real solution.
Drawing a number line, we can simplify that set of excluded x values to x ≤ -3 and 0 < x ≤ 3, which in interval notation is (-∞, -3] ∪ (0, 3].
The domain is the set of all x values which are not excluded, which is the opposite of that set (called the complement of that set). Looking at the number line, we can see that this is the set (-3, 0] ∪ (3, ∞).
If you have a graphing calculator, you can verify that this is a reasonable domain by looking at the values of x associated with points on the graph of y²(x² - 9) = 2x and the values of x that are not associated with any points. One graph of this equation is here: https://www.desmos.com/calculator/xmgi5whepv .
Range:
As we said above, the range of this equation is the set of all y values associated with solutions of this equation.
By going through the steps above, you may note that we might thus attempt to solve this equation for x in terms of y in order to see if there are any y values that cannot be associated with any x values.
Here, we would start by separating the variables: since x occurs on both sides of the equation, get all the x's on one side of the equation first: y²(x² - 9) - 2x = 0.
Simplify that multiplication so that we can eventually separate x from y: y²x² - 9y² - 2x = 0.
Rearrange this slightly and you can see that this a quadratic in x: y²x² - 2x - 9y² = 0.
We know everything there is to know about the solution set of quadratic equations.
In particular, there is no real solution if and only if the discriminant is negative.
The discriminant of this quadratic equation is (-2)² - 4y²(-9y²) which simplifies to 4 + 36y⁴.
Since y⁴ is never negative, this discriminant is also never negative (being a sum of nonnegative numbers), and thus we may conclude that the quadratic equation always has at least one real solution.
Thus, every value of y is associated with at least one value of x, so every value of y is in the range of the original equation.
Intercepts:
The intercepts are the points where the graph intersects the coordinate axes.
Let us see if there are any x-intercepts. All the points on the x-axis have a y-coordinate of 0, so if any solution is an x-intercept, it is associated with a y value of 0.
Thus, replacing y with 0 in the equation, we see that such solutions must solve the equation 0 = 2x. The only solution of this equation is x = 0, so the only x-intercept is the point (0,0) (some books want only the x coordinate, which would be 0).
Similarly, the y-intercept occurs when x is 0, so replacing x with 0, we see that a y intercept must solve the equation -9y² = 0, which is only solved by y = 0. Thus, (0,0) is the only x-intercept and the only y-intercept.
Symmetries:
A graph is symmetric about the x-axis if both (x, y) and (x, -y) solve the equation for all values of x.
Thus, let us replace y with -y and see if we get the same solution set for x.
Replacing y with -y in the original equation, we get (-y)²(x² - 9) = 2x.
This simplifies to y²(x² - 9) = 2x, which is identical to the original equation. Thus, -y is a solution of the equation whenever y is, for the same value of x, making the graph symmetric about the x-axis.
A graph is symmetric about the y-axis if, similarly, (x, y) and (-x, y) both solve the equation for every value of y.
Let's check by replacing x with -x: y²((-x²) - 9) = 2(-x) simplifies to y²(x² - 9) = -2x.
This is not the same equation and thus does not have the same solution set (this way of checking only works for polynomial equations). Thus, -x does not solve the equation whenever x does, and this graph is not symmetric about the y-axis.
A graph is symmetric about the origin (like a circle or regular polygon or star) if (-x, -y) is a solution whenever (x, y) is.
Let's check the same way: replacing x with -x and y with -y gives us the equation, after simplifying, y²(x² - 9) = -2x, which is not the same equation and thus does not have the same solution set. Thus, this graph is not symmetric about the origin.
Asymptotes:
A graph has a vertical asymptote x = k if the values of y approach either ∞ or -∞ as x approaches k from either side individually.
In an equation where we can solve for y as a function of x, where y is expressed as a fraction, like \(y = \pm\sqrt{\frac{2x}{x^2 - 9}}\), we may note that this behavior can only occur if the denominator is 0 while the numerator is non-zero.
We already noted that the denominator is 0 only when x = 3 or -3, and these values do not make the numerator 0, so x = 3 and x = -3 are the only vertical asymptotes of this graph.
Horizontal asymptotes y = k occur if y approaches k when x approaches ∞ or -∞.
Let's check: we know \(y = \pm\sqrt{\frac{2x}{x^2 - 9}}\).
As x approaches ∞, we intuitively note that the denominator will increase faster than the numerator, implying that the fraction will approach 0. Let's make that reasoning more reliable.
Multiply the top and bottom of this fraction by \(\frac{1}{x^2}\) to get the equivalent fraction \(\frac{2/x}{1 - (1/x^2)}\) (the square root stuff does not matter much here, since \(\sqrt{0} = -\sqrt{0} = 0\), but you can include it if you like).
As x approaches infinity, it is elementary to prove that \(\frac{1}{x}\) approaches 0.
Applying this fundamental fact to \(\frac{2}{x}\) and \(\frac{1}{x^2}\), which is \(\left(\frac{1}{x}\right)\left(\frac{1}{x}\right)\), we see that this fraction approaches \(\frac{0}{1 - 0}\), which is 0.
Thus, y = 0 is a horizontal asymptote of this function.
As x approaches -∞, we get the same result, so y = 0 is the only horizontal asymptote of this function.
Slant, or oblique, asymptotes occur if the graph approaches the line y = mx + b as x approaches ∞ or -∞.
Let's check: starting from \(y = \pm\sqrt{\frac{2x}{x^2 - 9}}\), we immediately see a problem: the square root function means that if this graph approaches any other graph as x approaches ∞, it would be something to do with \(\sqrt{x}\), not mx + b.
Thus, this graph has no slant asymptotes.
You may answer problem 6 by going through the same reasoning as above.
For problem 5, y²(x² - 9) = 2x, we can note the following:
Domain:
Unless otherwise mentioned, in elementary mathematics, y would be the range variable and x would be the domain variable.
The domain of this equation would thus be the set of all x values in the solution set of the equation (the solution set of this equation is the set of all ordered pairs, (x, y), that solve the equation).
We usually find this set by examining the operations involved in the equation, since most operations can be done with almost any number, except for a limited known set of numbers. Thus, our thinking is exclusionary in nature, not inclusionary.
In particular, both the left and right sides of this equation are polynomials. Polynomials only involve addition and multiplication, which is defined for all numbers, and thus do not exclude any values of x.
However, since this equation is not solved in a manner that allows us to easily see how y is related to x, we should then ask ourselves if there is any value of x that is not associated with any value of y (such a value of x would not be in the solution set of the equation).
In particular, notice that if we try to solve this equation using algebra for y as a function of x, we would have to divide both sides of the equation by the value of (x² - 9).
Division is an operation that is only defined for non-zero numbers. So if (x² - 9) is non-zero, we have no problems and we can solve for y as a function of whatever x is. However, we should inspect what happens if (x² - 9) is 0.
This occurs when x is either 3 or -3 (solve the equation x² - 9 = 0 for x). When x is either 3 or -3, the original equation becomes either 0 = 6 or 0 = -6, both of which are false statements. Thus, the values of 3 and -3 are not in the domain of this equation, since they are not associated with any solution of the equation.
Let us keep trying to solve for y as a function of x.
If x is neither 3 nor -3, then \(y^2 = \frac{2x}{x² - 9}\), and the Square Root Property then tells us that the only possible values of y are \(y = \pm\sqrt{\frac{2x}{x^2 - 9}}\).
Notice the principal square root function, √. In the Cartesian plane, where these equations are usually graphed, we only have real numbers. When the number inside the √ function is negative, the result is no longer a real number. When the number is nonnegative, we always get a real result.
Thus, if we are interested in domains of graphs in the Cartesian plane, we have to throw out any values of x that make the value of \(\frac{2x}{x^2 - 9}\) negative.
There are two ways that a fraction can be negative: the numerator is positive and the denominator is negative, or the denominator is negative and the numerator is positive.
Let us examine the first system of inequalities: 2x > 0 and x² - 9 < 0.
This system is solved when x > 0 and -3 < x < 3, which can be simplified to the single condition 0 < x < 3.
Thus, we know that all values of x in the interval where 0 < x < 3 are not in the domain of the equation.
Let us examine the second system of inequalities: 2x < 0 and x² - 9 > 0.
By similar algebra, we find the solution set of this system of inequalities is the interval x < -3.
Thus, having isolated y in terms of x, we see that when x = 3, x = -3, 0 < x < 3, or x < -3, the original equation has no real solution.
Drawing a number line, we can simplify that set of excluded x values to x ≤ -3 and 0 < x ≤ 3, which in interval notation is (-∞, -3] ∪ (0, 3].
The domain is the set of all x values which are not excluded, which is the opposite of that set (called the complement of that set). Looking at the number line, we can see that this is the set (-3, 0] ∪ (3, ∞).
If you have a graphing calculator, you can verify that this is a reasonable domain by looking at the values of x associated with points on the graph of y²(x² - 9) = 2x and the values of x that are not associated with any points. One graph of this equation is here: https://www.desmos.com/calculator/xmgi5whepv .
Range:
As we said above, the range of this equation is the set of all y values associated with solutions of this equation.
By going through the steps above, you may note that we might thus attempt to solve this equation for x in terms of y in order to see if there are any y values that cannot be associated with any x values.
Here, we would start by separating the variables: since x occurs on both sides of the equation, get all the x's on one side of the equation first: y²(x² - 9) - 2x = 0.
Simplify that multiplication so that we can eventually separate x from y: y²x² - 9y² - 2x = 0.
Rearrange this slightly and you can see that this a quadratic in x: y²x² - 2x - 9y² = 0.
We know everything there is to know about the solution set of quadratic equations.
In particular, there is no real solution if and only if the discriminant is negative.
The discriminant of this quadratic equation is (-2)² - 4y²(-9y²) which simplifies to 4 + 36y⁴.
Since y⁴ is never negative, this discriminant is also never negative (being a sum of nonnegative numbers), and thus we may conclude that the quadratic equation always has at least one real solution.
Thus, every value of y is associated with at least one value of x, so every value of y is in the range of the original equation.
Intercepts:
The intercepts are the points where the graph intersects the coordinate axes.
Let us see if there are any x-intercepts. All the points on the x-axis have a y-coordinate of 0, so if any solution is an x-intercept, it is associated with a y value of 0.
Thus, replacing y with 0 in the equation, we see that such solutions must solve the equation 0 = 2x. The only solution of this equation is x = 0, so the only x-intercept is the point (0,0) (some books want only the x coordinate, which would be 0).
Similarly, the y-intercept occurs when x is 0, so replacing x with 0, we see that a y intercept must solve the equation -9y² = 0, which is only solved by y = 0. Thus, (0,0) is the only x-intercept and the only y-intercept.
Symmetries:
A graph is symmetric about the x-axis if both (x, y) and (x, -y) solve the equation for all values of x.
Thus, let us replace y with -y and see if we get the same solution set for x.
Replacing y with -y in the original equation, we get (-y)²(x² - 9) = 2x.
This simplifies to y²(x² - 9) = 2x, which is identical to the original equation. Thus, -y is a solution of the equation whenever y is, for the same value of x, making the graph symmetric about the x-axis.
A graph is symmetric about the y-axis if, similarly, (x, y) and (-x, y) both solve the equation for every value of y.
Let's check by replacing x with -x: y²((-x²) - 9) = 2(-x) simplifies to y²(x² - 9) = -2x.
This is not the same equation and thus does not have the same solution set (this way of checking only works for polynomial equations). Thus, -x does not solve the equation whenever x does, and this graph is not symmetric about the y-axis.
A graph is symmetric about the origin (like a circle or regular polygon or star) if (-x, -y) is a solution whenever (x, y) is.
Let's check the same way: replacing x with -x and y with -y gives us the equation, after simplifying, y²(x² - 9) = -2x, which is not the same equation and thus does not have the same solution set. Thus, this graph is not symmetric about the origin.
Asymptotes:
A graph has a vertical asymptote x = k if the values of y approach either ∞ or -∞ as x approaches k from either side individually.
In an equation where we can solve for y as a function of x, where y is expressed as a fraction, like \(y = \pm\sqrt{\frac{2x}{x^2 - 9}}\), we may note that this behavior can only occur if the denominator is 0 while the numerator is non-zero.
We already noted that the denominator is 0 only when x = 3 or -3, and these values do not make the numerator 0, so x = 3 and x = -3 are the only vertical asymptotes of this graph.
Horizontal asymptotes y = k occur if y approaches k when x approaches ∞ or -∞.
Let's check: we know \(y = \pm\sqrt{\frac{2x}{x^2 - 9}}\).
As x approaches ∞, we intuitively note that the denominator will increase faster than the numerator, implying that the fraction will approach 0. Let's make that reasoning more reliable.
Multiply the top and bottom of this fraction by \(\frac{1}{x^2}\) to get the equivalent fraction \(\frac{2/x}{1 - (1/x^2)}\) (the square root stuff does not matter much here, since \(\sqrt{0} = -\sqrt{0} = 0\), but you can include it if you like).
As x approaches infinity, it is elementary to prove that \(\frac{1}{x}\) approaches 0.
Applying this fundamental fact to \(\frac{2}{x}\) and \(\frac{1}{x^2}\), which is \(\left(\frac{1}{x}\right)\left(\frac{1}{x}\right)\), we see that this fraction approaches \(\frac{0}{1 - 0}\), which is 0.
Thus, y = 0 is a horizontal asymptote of this function.
As x approaches -∞, we get the same result, so y = 0 is the only horizontal asymptote of this function.
Slant, or oblique, asymptotes occur if the graph approaches the line y = mx + b as x approaches ∞ or -∞.
Let's check: starting from \(y = \pm\sqrt{\frac{2x}{x^2 - 9}}\), we immediately see a problem: the square root function means that if this graph approaches any other graph as x approaches ∞, it would be something to do with \(\sqrt{x}\), not mx + b.
Thus, this graph has no slant asymptotes.
You may answer problem 6 by going through the same reasoning as above.