QUESTION IMAGE
Question
quadratic test
solve using the quadratic formula. $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$
- $y = 4x^2 - 8x + 3$
- $y = 9x^2 - 6x - 1$
graph quadratic function in vertex form. find a, h, k, vertex, axis of symmetry. create a table. graph the quadratic function.
- $y = 3(x + 2)^2 - 4$
- $y = -(x - 3)^2$
graph quadratic function in standard form. $ax^2 + bx + c$. find axis of symmetry using $x = \frac{-1(b)}{2(a)}$
- $y = 2x^2 + 4x - 5$
- $y = -x^2 + 6x + 5$
Problems 1 & 2: Solve using Quadratic Formula
Step1: Identify a,b,c for #1
For $y=4x^2-8x+3$, $a=4$, $b=-8$, $c=3$
Step2: Substitute into quadratic formula
$$x=\frac{-(-8)\pm\sqrt{(-8)^2-4(4)(3)}}{2(4)}$$
$$x=\frac{8\pm\sqrt{64-48}}{8}=\frac{8\pm\sqrt{16}}{8}=\frac{8\pm4}{8}$$
Step3: Calculate two roots for #1
$x=\frac{8+4}{8}=\frac{12}{8}=\frac{3}{2}$, $x=\frac{8-4}{8}=\frac{4}{8}=\frac{1}{2}$
Step4: Identify a,b,c for #2
For $y=9x^2-6x-1$, $a=9$, $b=-6$, $c=-1$
Step5: Substitute into quadratic formula
$$x=\frac{-(-6)\pm\sqrt{(-6)^2-4(9)(-1)}}{2(9)}$$
$$x=\frac{6\pm\sqrt{36+36}}{18}=\frac{6\pm\sqrt{72}}{18}=\frac{6\pm6\sqrt{2}}{18}$$
Step6: Simplify roots for #2
$x=\frac{6+6\sqrt{2}}{18}=\frac{1+\sqrt{2}}{3}$, $x=\frac{6-6\sqrt{2}}{18}=\frac{1-\sqrt{2}}{3}$
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Problems 3 & 4: Vertex Form Analysis & Graph Prep
Vertex form is $y=A(x-H)^2+K$, where:
- $A$ = vertical stretch/compression factor; negative = reflection over x-axis
- $(H,K)$ = vertex; axis of symmetry is $x=H$
For #3: $y=3(x+2)^2-4$ rewrite as $y=3(x-(-2))^2+(-4)$
For #4: $y=-(x-3)^2$ rewrite as $y=-1(x-3)^2+0$
Create tables by picking x-values around $x=H$, calculate corresponding y-values, then plot points and draw the parabola.
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Problems 5 & 6: Standard Form Axis of Symmetry
Step1: Identify a,b for #5
For $y=2x^2+4x-5$, $a=2$, $b=4$
Step2: Calculate axis of symmetry #5
$$x=\frac{-4}{2(2)}=\frac{-4}{4}=-1$$
Step3: Identify a,b for #6
For $y=-x^2+6x+5$, $a=-1$, $b=6$
Step4: Calculate axis of symmetry #6
$$x=\frac{-6}{2(-1)}=\frac{-6}{-2}=3$$
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- $x=\frac{3}{2}$ and $x=\frac{1}{2}$
- $x=\frac{1+\sqrt{2}}{3}$ and $x=\frac{1-\sqrt{2}}{3}$
- $A=3$, $H=-2$, $K=-4$; Vertex: $(-2,-4)$; Axis of Symmetry: $x=-2$
Sample table:
| x | y |
|---|---|
| -3 | 5 |
| -2 | -4 |
| -1 | -1 |
| 0 | 8 |
- $A=-1$, $H=3$, $K=0$; Vertex: $(3,0)$; Axis of Symmetry: $x=3$
Sample table:
| x | y |
|---|---|
| 2 | -1 |
| 3 | 0 |
| 4 | -1 |
| 5 | -4 |
- Axis of Symmetry: $x=-1$
- Axis of Symmetry: $x=3$