9.5: Graphing Quadratic Equations (2024)

Example \(\PageIndex{19}\)

Graph \(y=x2−6x+8\).

Answer

Example \(\PageIndex{20}\)

Graph the parabola \(y=x^2+2x−8\).

Answer

y:(0,−8); x:(2,0),(−4,0);
axis: x=−1; vertex: (−1,−9);

Example \(\PageIndex{21}\)

Graph the parabola \(y=x^2−8x+12\).

Answer

y:(0,12); x:(2,0),(6,0);
axis: x=4; vertex:(4,−4);

Definition: GRAPH A QUADRATIC EQUATION IN TWO VARIABLES.

1. Write the quadratic equation with yy on one side.
2. Determine whether the parabola opens upward or downward.
3. Find the axis of symmetry.
4. Find the vertex.
5. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
6. Find the x-intercepts.
7. Graph the parabola.

Example \(\PageIndex{22}\)

Graph \(y=−x^2+6x−9\).

Answer

The equation y has on one side.
Since a is −1, the parabola opens downward. To find the axis of symmetry, find \(x=−\frac{b}{2a}\).		The axis of symmetry is x=3. The vertex is on the line x=3.
Find y when x=3.		The vertex is (3,0).
The y-intercept occurs when x=0. Substitute x=0. Simplify. The point (0,−9) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is (6,−9). Point symmetric to the y-intercept is (6,−9)		(0,−9).
The x-intercept occurs when y=0.
Substitute y=0.
Factor the GCF.
Factor the trinomial.
Solve for x.
Connect the points to graph the parabola.

Example \(\PageIndex{23}\)

Graph the parabola \(y=−3x^2+12x−12\).

Answer

y:(0,−12); x:(2,0);
axis: x=2; vertex:(2,0);

Example \(\PageIndex{24}\)

Graph the parabola \(y=25x^2+10x+1\).

Answer

y:(0,1); x:(−15,0);
axis: x=−15; vertex:(−15,0);

Example \(\PageIndex{25}\)

Graph \(y=x^2+4x+5\).

Answer

The equation has y on one side.
Since a is 1, the parabola opens upward.
\(x=−\frac{b}{2a}\).		x=−2.
The vertex is on the line x=−2.
Find y when x=−2.		(−2,1).
The y-intercept occurs when x=0. Substitute x=0. Simplify. The point (0,5) is two units to the right of the line of symmetry. The point two units to the left of the line of symmetry is (−4,5).		(0,5). (−4,5)
The x- intercept occurs when y=0.
Substitute y=0. Test the discriminant.
		\(b^2−4ac\) \(42−4⋅15\) \(16−20\) \(−4\)
Since the value of the discriminant is negative, there is no solution and so no x- intercept. Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.

Example \(\PageIndex{26}\)

Graph the parabola \(y=2x^2−6x+5\).

Answer

y:(0,5); x:none;
axis: \(x=\frac{3}{2}\); vertex:\((\frac{3}{2},\frac{1}{2})\);

Example \(\PageIndex{27}\)

Graph the parabola \(y=−2x^2−1\).

Answer

y:(0,−1); x:none;
axis: x=0; vertex:(0,−1);

Example \(\PageIndex{28}\)

Graph \(y=2x^2−4x−3\).

Answer

The equation y has one side. Since a is 2, the parabola opens upward.
To find the axis of symmetry, find\(x=−\frac{b}{2a}\)		The vertex is x=1
The vertex on the line x=1.
Find y when x=1		(1,−5)
The y-intercept occurs when x=0.
Substitute x=0.
Simplify.		The y-intercept is (0,−3)
The point (0,−3) is one unit to the left of the line of symmetry. The point one unit to the right of the line of symmetry is (2,−3)		Point symmetric to the y-intercept is (2,−3).
The x-intercept occurs when y=0
Substitute y=0
Use the Quadratic Formula.
Substitute in the values of a, b, c.
Simplify.
Simplify inside the radical.
Simplify the radical.
Factor the GCF.
Remove common factors.
Write as two equations.
Approximate the values.
		The approximate values of the x-intercepts are (2.5,0) and (−0.6,0).
Graph the parabola using the points found.

Example \(\PageIndex{29}\)

Graph the parabola \(y=5x^2+10x+3\).

Answer

y:(0,3); x:(−1.6,0),(−0.4,0);
axis: x=−1; vertex:(−1,−2);

Example \(\PageIndex{30}\)

Graph the parabola \(y=−3x^2−6x+5\).

Answer

y:(0,5); x:(0.6,0),(−2.6,0);
axis: x=−1; vertex:(−1,8);

Definition: MINIMUM OR MAXIMUM VALUES OF A QUADRATIC EQUATION

The y-coordinate of the vertex of the graph of a quadratic equation is the

• minimum value of the quadratic equation if the parabola opens upward.
• maximum value of the quadratic equation if the parabola opens downward.

Example \(\PageIndex{31}\)

Find the minimum value of the quadratic equation \(y=x^2+2x−8\).

Answer

Since a is positive, the parabola opens upward.
The quadratic equation has a minimum.
Find the axis of symmetry.		x=−1
The vertex is on the line x=−1.
Find y when x=−1.		(−1,−9)
Since the parabola has a minimum, the y-coordinate of the vertex is the minimum y-value of the quadratic equation.
The minimum value of the quadratic is −9 and it occurs when x=−1.
Show the graph to verify the result.

Example \(\PageIndex{32}\)

Find the maximum or minimum value of the quadratic equation \(y=x^2−8x+12\).

Answer

The minimum value is −4 when x=4.

Example \(\PageIndex{33}\)

Find the maximum or minimum value of the quadratic equation \(y=−4x^2+16x−11\).

Answer

The maximum value is 5 when x=2.

Example \(\PageIndex{34}\)

The quadratic equation \(h=−16t^2+v_{0}t+h_{0}\) models the height of a volleyball hit straight upwards with velocity 176 feet per second from a height of 4 feet.

1. How many seconds will it take the volleyball to reach its maximum height?
2. Find the maximum height of the volleyball.

Answer

\(h=−16t^2+176t+4\)

Since a is negative, the parabola opens downward.

The quadratic equation has a maximum.

1.
\[\begin{array} {ll} {}&{t=−\frac{b}{2a}}\\ {\text{Find the axis of symmetry.}}& {t=−\frac{176}{2(−16)}}\\ {}&{t=5.5}\\ {}&{\text{The axis of symmetry is} t = 5.5}\\ {\text{The vertex is on the line} t=5.5}& {\text{The maximum occurs when} t =5.5 \text{seconds.}}\\ \nonumber \end{array}\]

2.

Find h when t=5.5.
Use a calculator to simplify.
		The vertex is (5.5,488)
Since the parabola has a maximum, the h-coordinate of the vertex is the maximum y-value of the quadratic equation.		The maximum value of the quadratic is 488 feet and it occurs when t=5.5 seconds.