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Section8.6Additional Practice Related to Parabolas

Subsection8.6.1Exercises

For each equation, state the vertex, axis of symmetry, concavity, and all intercepts of the parabola.

1

\(y=x^2+4x-12\)

Solution

The vertex of the parabola is \((-2,16)\text{.}\) The axis of symmetry is the line with equation \(x=-2\text{.}\) The parabola is concave up. The \(x\)-intercepts are \((-6,0)\) and \((2,0)\text{.}\) The \(y\)-intercept is \((0,-12)\text{.}\)

2

\(y=-x^2-3\)

Solution

The vertex of the parabola is \((0,-3)\text{.}\) The axis of symmetry is the line with equation \(x=0\) (i.e., the \(y\)-axis). The parabola is concave down. The \(y\)-intercept is \((0,-3)\text{.}\) The parabola has no \(x\)-intercepts.

3

\(y=4x^2+8x\)

Solution

The vertex of the parabola is \((-1,-4)\text{.}\) The axis of symmetry is the line with equation \(x=-1\text{.}\) The parabola is concave up. The \(x\)-intercepts are \((-2,0)\) and \((0,0)\text{.}\) The \(y\)-intercept is \((0,0)\text{.}\)

4

\(y=\frac{1}{2}x^2-4x+13\)

Solution

The vertex of the parabola is \((4,5)\text{.}\) The axis of symmetry is the line with equation \(x=4\text{.}\) The parabola is concave up. The \(y\)-intercept is \((0,13)\text{.}\) The parabola has no \(x\)-intercepts.

5

\(y=-2x^2+7x-4\)

Solution

The vertex of the parabola is \(\left(\frac{7}{4},\frac{17}{8}\right)\text{.}\) The axis of symmetry is the line with equation \(x=\frac{7}{4}\text{.}\) The parabola is concave down. The \(x\)-intercepts are \(\left(\frac{7-\sqrt{17}}{4},0\right)\) and \(\left(\frac{7+\sqrt{17}}{4},0\right)\text{.}\) The \(y\)-intercept is \((0,-4)\text{.}\)

An out-of-the-ox question for you.

6

The points given in Table 8.6.1 all lie on the same parabola. Determine the axis-of-symmetry for the parabola and complete the missing entries in the table.

\(x\) \(y\)
\(-3\) \(-28\)
\(-2\) \(\)
\(-1\) \(\)
\(0\) \(8\)
\(1\) \(\)
\(2\) \(12\)
\(3\) \(8\)
\(4\) \(0\)
\(5\) \(-12\)
\(6\) \(\)
Table8.6.1Points from a parabola

Solution

The axis of symmetry is the line with equation \(x=1.5\text{.}\)

\(x\) \(y\)
\(-3\) \(-28\)
\(-2\) \(-12\)
\(-1\) \(0\)
\(0\) \(8\)
\(1\) \(12\)
\(2\) \(12\)
\(3\) \(8\)
\(4\) \(0\)
\(5\) \(-12\)
\(6\) \(-28\)
Table8.6.2Points from a parabola

For each equation, state the vertex, axis of symmetry, concavity, and all intercepts of the parabola.

7

\(y=-(x+4)^2+9\)

Solution

The vertex of the parabola is \((-4,9)\text{.}\) The axis of symmetry is the line with equation \(x=-4\text{.}\) The parabola is concave down. The \(x\)-intercepts are \((-1,0)\) and \((-7,0)\text{.}\) The \(y\)-intercept is \((0,9)\text{.}\)

8

\(y=3x^2-12\)

Solution

The vertex of the parabola is \((0,-12)\text{.}\) The axis of symmetry is the line with equation \(x=0\) (i.e. the \(y\)-axis). The parabola is concave up. The \(x\)-intercepts are \((-2,0)\) and \((2,0)\text{.}\) The \(y\)-intercept is \((0,-12)\text{.}\)

9

\(y=2(x-5)^2+16\)

Solution

The vertex of the parabola is \((5,16)\text{.}\) The axis of symmetry is the line with equation \(x=5\text{.}\) The parabola is concave up. The \(y\)-intercept is \((0,16)\text{.}\) The parabola has no \(x\)-intercepts.

10

\(y=-2(x-7)+40\)

Solution

The vertex of the parabola is \((7,40)\text{.}\) The axis of symmetry is the line with equation \(x=7\text{.}\) The parabola is concave down. The \(x\)-intercepts are \((7+2\sqrt{5},0)\) and \((7-2\sqrt{5},0)\text{.}\) The \(y\)-intercept is \((0,40)\text{.}\)

11

\(y=-6(x+4)^2\)

Solution

The vertex of the parabola is \((-4,0)\text{.}\) The axis of symmetry is the line with equation \(x=-4\text{.}\) The parabola is concave down. The only \(x\)-intercept is \((-4,0)\text{.}\) The \(y\)-intercept is \((0,-96)\text{.}\)

For each graph, determine an equation that would produce the parabola. State the equation in both graphing form and standard form.

12

The graph of a parabola that opens concave down from the vertex \((-2,6)\text{.}\)  The parabola also passes through the point \((0,-2)\text{.}\)
Figure8.6.3Determine an equation for the parabola

Solution

The graphing form of the equation of the parabola shown in Figure 8.6.3 is \(y=-2(x+2)^2+6\) and the standard form of the equation is \(y=-2x^2-8x-2\text{.}\)

13

The graph of a parabola that opens concave up from the vertex \((0,-3)\text{.}\)  The parabola also passes through the point \((2,3)\text{.}\)
Figure8.6.4Determine an equation for the parabola

Solution

Both the graphing form and standard form of the equation of the parabola shown in Figure 8.6.4 is \(y=1.5x^2-3\text{.}\)

14

The graph of a parabola that opens concave up from the point \((4,0)\text{.}\)  The parabola also passes through the point \((0,4)\text{.}\)
Figure8.6.5Determine an equation for the parabola

Solution

The graphing form of the equation of the parabola shown in Figure 8.6.5 is \(y=\frac{1}{4}(x-4)^2\) and the standard form of the equation is \(y=\frac{1}{4}x^2-2x+4\text{.}\)

Determine the vertex of each parabola after first completing the square to determine the graphing form of the equation.

15

\(y=x^2-8x+7\)

Solution

The graphing form of the equation is \(y=(x-4)^2-9\text{.}\) The vertex of the parabola is \((4,-9)\text{.}\)

16

\(y=x^2-5x-2\)

Solution

The graphing form of the equation is \(y=\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\text{.}\) The vertex of the parabola is \(\left(\frac{5}{2},-\frac{33}{4}\right)\text{.}\)

17

\(y=2x^2+12x+20\)

Solution

The graphing form of the equation is \(y=2(x+3)^2+2\text{.}\) The vertex of the parabola is \((-3,2)\text{.}\)

18

\(y=-3x^2-24x-15\)

Solution

The graphing form of the equation is \(y=-3(x+4)^2+33\text{.}\) The vertex of the parabola is \((-4,33)\text{.}\)

19

\(y=-2x^2+11x\)

Solution

The graphing form of the equation is \(y=-2\left(x-\frac{11}{4}\right)^2+\frac{121}{8}\text{.}\) The vertex of the parabola is \(\left(\frac{11}{4},\frac{121}{8}\right)\text{.}\)

20

\(y=-x^2+10x-28\)

Solution

The graphing form of the equation is \(y=-(x-5)^2-3\text{.}\) The vertex of the parabola is \((5,-3)\text{.}\)