Question

forms of quadratics
name: _______
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. $y = 3x^2 - 6x + 4$

Explanation:

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Part 1: Classify quadratic forms

Step1: Recall standard/vertex form rules

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

Step2: Classify each function

1. $y=x^2-2x-35$ matches $y=ax^2+bx+c$ → Standard form
2. $y=3(x-7)^2+3$ matches $y=a(x-h)^2+k$ → Vertex form
3. $y=-\frac{2}{3}(x-4)^2+7$ matches $y=a(x-h)^2+k$ → Vertex form
4. $y=-2x^2+16x-24$ matches $y=ax^2+bx+c$ → Standard form

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Part 2: Vertex from vertex form

Step1: Use vertex form vertex rule

For $y=a(x-h)^2+k$, vertex is $(h,k)$

Step2: Calculate each vertex

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

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Part 3: Vertex from standard form

Step1: Find x-coordinate of vertex

Use $x=-\frac{b}{2a}$ for $y=ax^2+bx+c$

Step2: Find y-coordinate via substitution

Substitute $x$ into the original function to find $y$

Step3: Solve each function

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

• $a=2, b=-16$
• $x=-\frac{-16}{2(2)} = \frac{16}{4}=4$
• $y=2(4)^2-16(4)+31=32-64+31=-1$ → $(4, -1)$

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

• $a=-1, b=-4$
• $x=-\frac{-4}{2(-1)} = \frac{4}{-2}=-2$
• $y=-(-2)^2-4(-2)+1=-4+8+1=5$ → $(-2, 5)$

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

• $a=3, b=-6$
• $x=-\frac{-6}{2(3)} = \frac{6}{6}=1$
• $y=3(1)^2-6(1)+4=3-6+4=1$ → $(1, 1)$

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Answer:

Part 1: Form Classification

1. Standard form
2. Vertex form
3. Vertex form
4. Standard form

Part 2: Vertex from Vertex Form

1. $(7, -1)$
2. $(-2, -5)$
3. $(3, 0)$
4. $(2, 4)$
5. $(-1, -3)$
6. $(-4, 0)$
7. $(5, 1)$
8. $(-6, 10)$

Part 3: Vertex from Standard Form

1. $(4, -1)$
2. $(-2, 5)$
3. $(1, 1)$