Skip to main content

Section 8.6 Additional Practice Related to Parabolas

Exercises Exercises

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.

\(x\) \(y\)
\(-3\) \(-28\)
\(-2\) \(\)
\(-1\) \(\)
\(0\) \(8\)
\(1\) \(\)
\(2\) \(12\)
\(3\) \(8\)
\(4\) \(0\)
\(5\) \(-12\)
\(6\) \(\)
Table 8.6.1. Points from a parabola

Solution

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

\(x\) \(y\)
\(-3\) \(-28\)
\(-2\) \(-12\)
\(-1\) \(0\)
\(0\) \(8\)
\(1\) \(12\)
\(2\) \(12\)
\(3\) \(8\)
\(4\) \(0\)
\(5\) \(-12\)
\(6\) \(-28\)
Table 8.6.2. Points 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 \((0,-2)\text{.}\)
Figure 8.6.3. Determine 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 \((2,3)\text{.}\)
Figure 8.6.4. Determine 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 \((0,4)\text{.}\)
Figure 8.6.5. Determine 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 \((5,-3)\text{.}\)