Question

forms of quadratics
tell whether the quadratic function is in standard form or vertex form.

1. $y = x^2 - 2x - 35$
2. $y = 3(x-1)^2 + 3$
3. $y = \frac{2}{3}(x-4)^2 + 7$
4. $y = -2x^2 + 16x - 24$

identify the vertex of the quadratic function in vertex form.

1. $y = 3(x-7)^2 - 1$
2. $y = 3(x+2)^2 - 5$
3. $y = (x-3)^2$
4. $y = -4(x-2)^2 + 4$
5. $y = 2(x+1)^2 - 3$
6. $y = (x+4)^2$
7. $y = \frac{1}{2}(x-5)^2 + 1$
8. $y = -(x+6)^2 + 10$

identify the vertex of the quadratic function in standard form. remember to use $x = \frac{-b}{2a}$

1. $y = 2x^2 - 16x + 31$
2. $y = -x^2 - 4x + 1$
3. $y = 3x^2 - 6x + 4$

given a quadratic equation in vertex form, find the vertex, axis of symmetry, whether the graph opens up or down, the maximum or minimum, and the y-intercept. graph it!

1. $y = -2(x+2)^2 + 4$

vertex: ________
axis of symmetry: ________
opens: up down
maximum minimum
max/min value: ________
y-intercept: ________

Explanation:

Step1: Classify quadratic forms

Standard form: $y=ax^2+bx+c$; Vertex form: $y=a(x-h)^2+k$

1. $y=x^2-2x-35$: Standard form
2. $y=3(x-1)^2+3$: Vertex form
3. $y=-\frac{2}{3}(x-4)^2+7$: Vertex form
4. $y=-2x^2+16x-24$: Standard form

Step2: Find vertices (vertex form)

Vertex of $y=a(x-h)^2+k$ is $(h,k)$

1. $y=3(x-7)^2-1$: $(7, -1)$
2. $y=3(x+2)^2-5=3(x-(-2))^2-5$: $(-2, -5)$
3. $y=(x-3)^2=(x-3)^2+0$: $(3, 0)$
4. $y=-4(x-2)^2+4$: $(2, 4)$
5. $y=2(x+1)^2-3=2(x-(-1))^2-3$: $(-1, -3)$
6. $y=(x+4)^2=(x-(-4))^2+0$: $(-4, 0)$
7. $y=\frac{1}{2}(x-5)^2+1$: $(5, 1)$
8. $y=-(x+6)^2+10=-(x-(-6))^2+10$: $(-6, 10)$

Step3: Find vertices (standard form)

First calculate $x=-\frac{b}{2a}$, then find $y$

1. $y=2x^2-16x+31$:

$x=-\frac{-16}{2\times2}=4$
$y=2(4)^2-16(4)+31=32-64+31=-1$
Vertex: $(4, -1)$

1. $y=-x^2-4x+1$:

$x=-\frac{-4}{2\times(-1)}=-2$
$y=-(-2)^2-4(-2)+1=-4+8+1=5$
Vertex: $(-2, 5)$

1. $y=3x^2-6x+4$:

$x=-\frac{-6}{2\times3}=1$
$y=3(1)^2-6(1)+4=3-6+4=1$
Vertex: $(1, 1)$

Step4: Analyze $y=-2(x+2)^2+4$

Step4a: Find vertex

Rewrite as $y=-2(x-(-2))^2+4$, vertex: $(-2, 4)$

Step4b: Axis of symmetry

Axis is $x=h$, so $x=-2$

Step4c: Direction of opening

$a=-2<0$, so opens down

Step4d: Max/min value

Opens down, so maximum at $y=4$

Step4e: Find y-intercept

Set $x=0$:
$y=-2(0+2)^2+4=-2(4)+4=-8+4=-4$
Y-intercept: $(0, -4)$

Answer:

1. Standard form
2. Vertex form
3. Vertex form
4. Standard form
5. $(7, -1)$
6. $(-2, -5)$
7. $(3, 0)$
8. $(2, 4)$
9. $(-1, -3)$
10. $(-4, 0)$
11. $(5, 1)$
12. $(-6, 10)$
13. $(4, -1)$
14. $(-2, 5)$
15. $(1, 1)$

16.
Vertex: $(-2, 4)$
Axis of symmetry: $x=-2$
Opens: down
Maximum
Max/Min Value: $4$
y-intercept: $(0, -4)$
(Graph: Plot vertex $(-2,4)$, y-intercept $(0,-4)$, mirror the y-intercept over $x=-2$ to get $(-4,-4)$, then draw a downward-opening parabola through these points)