Section 10.6 Additional Practice Related to Parabolas
¶Exercises Exercises
For each equation, state the vertex, axis of symmetry, concavity, and all intercepts of the parabola.
1.
\(y=x^2+4x-12\)
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.
\(y=-x^2-3\)
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.
\(y=4x^2+8x\)
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.
\(y=\frac{1}{2}x^2-4x+13\)
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.
\(y=-2x^2+7x-4\)
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 Figure 10.6.1 all lie on the same parabola. Determine the axis-of-symmetry for the parabola and complete the missing entries in the table.
\(x\)
\(y\)
\(-3\)
\(-28\)
\(-2\)
\(\)
\(-1\)
\(\)
\(0\)
\(8\)
\(1\)
\(\)
\(2\)
\(12\)
\(3\)
\(8\)
\(4\)
\(0\)
\(5\)
\(-12\)
\(6\)
\(\)
The axis of symmetry is the line with equation \(x=1.5\text{.}\)
\(x\)
\(y\)
\(-3\)
\(-28\)
\(-2\)
\(-12\)
\(-1\)
\(0\)
\(0\)
\(8\)
\(1\)
\(12\)
\(2\)
\(12\)
\(3\)
\(8\)
\(4\)
\(0\)
\(5\)
\(-12\)
\(6\)
\(-28\)
For each equation, state the vertex, axis of symmetry, concavity, and all intercepts of the parabola.
7.
\(y=-(x+4)^2+9\)
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.
\(y=3x^2-12\)
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.
\(y=2(x-5)^2+16\)
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.
\(y=-2(x-7)+40\)
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.
\(y=-6(x+4)^2\)
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 graphing form of the equation of the parabola shown in Figure 10.6.3 is \(y=-2(x+2)^2+6\) and the standard form of the equation is \(y=-2x^2-8x-2\text{.}\)
13.
Both the graphing form and standard form of the equation of the parabola shown in Figure 10.6.4 is \(y=1.5x^2-3\text{.}\)
14.
The graphing form of the equation of the parabola shown in Figure 10.6.5 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.
\(y=x^2-8x+7\)
The graphing form of the equation is \(y=(x-4)^2-9\text{.}\) The vertex of the parabola is \((4,-9)\text{.}\)
16.
\(y=x^2-5x-2\)
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.
\(y=2x^2+12x+20\)
The graphing form of the equation is \(y=2(x+3)^2+2\text{.}\) The vertex of the parabola is \((-3,2)\text{.}\)
18.
\(y=-3x^2-24x-15\)
The graphing form of the equation is \(y=-3(x+4)^2+33\text{.}\) The vertex of the parabola is \((-4,33)\text{.}\)
19.
\(y=-2x^2+11x\)
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.
\(y=-x^2+10x-28\)
The graphing form of the equation is \(y=-(x-5)^2-3\text{.}\) The vertex of the parabola is \((5,-3)\text{.}\)