10.5 Graphing quadratic equations (Page 3/15)

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Do these results agree with our graphs? See [link] .

This figure shows an two graphs side by side. The graph on the left side shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The vertex is at the point (-2, -1). Three points are plotted on the curve at (-3, 0), (-1, 0), and (0, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -2. Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the statement “y-intercept (0, 3)”. Below that is the statement “x-intercepts (-1, 0) and (-3, 0)”. The graph on the right side shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The vertex is at the point (2, 7). Three points are plotted on the curve at (-0.6, 0), (4.6, 0), and (0, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2. Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the statement “y-intercept (0, 3)”. Below that is the statement “x-intercepts (2 plus square root of 7, 0) is approximately equal to (4.6, 0) and (2 minus square root of 7, 0) is approximately equal to (-0.6, 0).”

Find the intercepts of a parabola.

To find the intercepts of a parabola with equation $y = a x^{2} + b x + c$ :

\begin{matrix} y -intercept & x -intercepts \\ Let x = 0 and solve for y . & Let y = 0 and solve for x . \end{matrix}

Find the intercepts of the parabola $y = x^{2} - 2 x - 8$ .

Solution


To find the y -intercept, let $x = 0$ and solve for y .
	When $x = 0$ , then $y = −8$ . The y -intercept is the point $(0, −8)$ .

To find the x -intercept, let $y = 0$ and solve for x .
Solve by factoring.

When $y = 0$ , then $x = 4 or x = −2$ . The x -intercepts are the points $(4, 0)$ and $(−2, 0)$ .

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Find the intercepts of the parabola $y = x^{2} + 2 x - 8 .$

$y : (0, −8); x : (−4, 0), (2, 0)$

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Find the intercepts of the parabola $y = x^{2} - 4 x - 12 .$

$y : (0, −12); x : (6, 0), (−2, 0)$

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In this chapter, we have been solving quadratic equations of the form $a x^{2} + b x + c = 0$ . We solved for $x$ and the results were the solutions to the equation.

We are now looking at quadratic equations in two variables of the form $y = a x^{2} + b x + c$ . The graphs of these equations are parabolas. The x -intercepts of the parabolas occur where $y = 0$ .

For example:

\begin{matrix} Quadratic equation & Quadratic equation in two variables \\ y = x^{2} - 2 x - 15 \\ \begin{matrix} x^{2} - 2 x - 15 & = & 0 \\ (x - 5) (x + 3) & = & 0 \end{matrix} & let y = 0 \begin{matrix} 0 = x^{2} - 2 x - 15 \\ 0 = (x - 5) (x + 3) \end{matrix} \\ \begin{matrix} x - 5 & = & 0 & x + 3 & = & 0 \\ x & = & 5 & x & = & −3 \end{matrix} & \begin{matrix} x - 5 & = & 0 & x + 3 & = & 0 \\ x & = & 5 & x & = & −3 \end{matrix} \\ (5, 0) and (−3, 0) \\ x -intercepts \end{matrix}

The solutions of the quadratic equation are the $x$ values of the x -intercepts.

Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the equations give the x -intercepts of the graphs, the number of x -intercepts is the same as the number of solutions.

Previously, we used the discriminant to determine the number of solutions of a quadratic equation of the form $a x^{2} + b x + c = 0$ . Now, we can use the discriminant to tell us how many x -intercepts there are on the graph.

This figure shows three graphs side by side. The leftmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the lower right quadrant. Below the graph is the inequality b squared minus 4 a c greater than 0. Below that is the statement “Two solutions”. Below that is the statement “ Two x-intercepts”. The middle graph shows an downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is on the x-axis. Below the graph is the equation b squared minus 4 a c equals 0. Below that is the statement “One solution”. Below that is the statement “ One x-intercept”. The rightmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the upper left quadrant. Below the graph is the inequality b squared minus 4 a c less than 0. Below that is the statement “No real solutions”. Below that is the statement “ No x-intercept”.

Before you start solving the quadratic equation to find the values of the x -intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

Find the intercepts of the parabola $y = 5 x^{2} + x + 4$ .

Solution


To find the y -intercept, let $x = 0$ and solve for y .	When $x = 0$ , then $y = 4$ . The y -intercept is the point $(0, 4)$ .

To find the x -intercept, let $y = 0$ and solve for x .
Find the value of the discriminant to predict the number of solutions and so x -intercepts.	$\begin{matrix} \begin{matrix} b^{2} & - & 4 a c \\ 1^{2} & - & 4 \cdot 5 \cdot 4 \\ 1 & - & 80 \end{matrix} \\ −79 \end{matrix}$
Since the value of the discriminant is negative, there is no real solution to the equation.	There are no x -intercepts.

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Find the intercepts of the parabola $y = 3 x^{2} + 4 x + 4 .$

$y : (0, 4); x : none$

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Find the intercepts of the parabola $y = x^{2} - 4 x - 5 .$

$y : (0, −5); x : (5, 0) (−1, 0)$

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Find the intercepts of the parabola $y = 4 x^{2} - 12 x + 9$ .

Solution


To find the y -intercept, let $x = 0$ and solve for y .
	When $x = 0$ , then $y = 9$ . The y -intercept is the point $(0, 9)$ .

To find the x -intercept, let $y = 0$ and solve for x .
Find the value of the discriminant to predict the number of solutions and so x -intercepts.	$\begin{matrix} \begin{matrix} b^{2} & - & 4 a c \\ 2^{2} & - & 4 \cdot 4 \cdot 9 \\ 144 & - & 144 \end{matrix} \\ 0 \end{matrix}$
	Since the value of the discriminant is 0, there is no real solution to the equation. So there is one x -intercept.
Solve the equation by factoring the perfect square trinomial.
Use the Zero Product Property.
Solve for x .
	When $y = 0$ , then $\frac{3}{2} = x .$
	The x -intercept is the point $(\frac{3}{2}, 0) .$

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Find the intercepts of the parabola $y = − x^{2} - 12 x - 36 .$

$y : (0, −36); x : (−6, 0)$

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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