Algebra 2 Trig1 Graphing Quadratic Equations Answer Key - Fill ... - Free Printable

Let’s solve each problem step by step. We’ll go one at a time, carefully checking our work.

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Problem 4: y = -1·3x²

First, simplify the expression:
y = -3x²

This is in standard form: y = ax² + bx + c
Here, a = -3, b = 0, c = 0

Since a is negative, the graph opens down.

Vertex? For any quadratic y = ax² + bx + c, vertex x-coordinate is at x = -b/(2a)
→ x = -0 / (2*(-3)) = 0
Plug x=0 into equation: y = -3*(0)² = 0 → Vertex is (0, 0)

Axis of symmetry: vertical line through vertex → x = 0

✔ Final for #4:
Standard form: y = -3x²
Opens: down
Vertex: (0, 0)
Axis of symmetry: x = 0

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Problem 5: y = -x + 9x²

Rewrite in standard form (descending powers):
y = 9x² - x

So, a = 9, b = -1, c = 0

Since a is positive, graph opens up

Vertex x = -b/(2a) = -(-1)/(2*9) = 1/18

Now find y: plug x = 1/18 into y = 9x² - x
y = 9*(1/18)² - (1/18)
= 9*(1/324) - 1/18
= 9/324 - 1/18
Simplify 9/324 = 1/36
So y = 1/36 - 1/18 = 1/36 - 2/36 = -1/36

Vertex: (1/18, -1/36)

Axis of symmetry: x = 1/18

✔ Final for #5:
Standard form: y = 9x² - x
Opens: up
Vertex: (1/18, -1/36)
Axis of symmetry: x = 1/18

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Problem 6: y = 3x² + 5x³ + 2

Wait — this has an x³ term! That means it’s not a quadratic. Quadratics only have up to x².

But the worksheet says “Write the quadratic equation...” — so maybe there’s a typo? Let’s check again.

Original: y = 3x² + 5x³ + 2 → that’s cubic, not quadratic.

Perhaps it was meant to be y = 3x² + 5x + 2? That would make sense.

Looking at the context — all other problems are quadratics. Also, problem 7,8,9 are clearly quadratics. So likely a typo: probably meant y = 3x² + 5x + 2

We’ll assume that’s what was intended.

So: y = 3x² + 5x + 2 → already in standard form
a = 3, b = 5, c = 2 → a > 0 → opens up

Vertex x = -b/(2a) = -5/(2*3) = -5/6

Find y: plug x = -5/6 into y = 3x² + 5x + 2

Compute step by step:

x² = (-5/6)² = 25/36
3x² = 3 * 25/36 = 75/36 = 25/12

5x = 5 * (-5/6) = -25/6

Now add:
y = 25/12 - 25/6 + 2

Convert to twelfths:
25/12 - 50/12 + 24/12 = (25 - 50 + 24)/12 = (-1)/12

So vertex: (-5/6, -1/12)

Axis of symmetry: x = -5/6

✔ Final for #6 (assuming typo corrected):
Standard form: y = 3x² + 5x + 2
Opens: up
Vertex: (-5/6, -1/12)
Axis of symmetry: x = -5/6

*(Note: If the original was truly y = 5x³ + 3x² + 2, then it’s not a quadratic and can’t be solved as such. But given context, we assume typo.)*

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Problem 7: y = 2x² + 4x - 1

Already in standard form: a=2, b=4, c=-1 → a>0 → opens up

Axis of symmetry: x = -b/(2a) = -4/(4) = -1

Vertex: plug x=-1 into equation
y = 2(-1)² + 4(-1) -1 = 2(1) -4 -1 = 2 -4 -1 = -3
Vertex: (-1, -3)

✔ Final for #7:
Axis of symmetry: x = -1
Vertex: (-1, -3)

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Problem 8: y = x² - 2x + 5

a=1, b=-2, c=5 → opens up

Axis of symmetry: x = -(-2)/(2*1) = 2/2 = 1

Vertex: plug x=1 → y = (1)² -2(1) +5 = 1 -2 +5 = 4
Vertex: (1, 4)

✔ Final for #8:
Axis of symmetry: x = 1
Vertex: (1, 4)

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Problem 9: y = 3x² - 3

a=3, b=0, c=-3 → opens up

Axis of symmetry: x = -0/(2*3) = 0

Vertex: plug x=0 → y = 3(0) -3 = -3 → (0, -3)

✔ Final for #9:
Axis of symmetry: x = 0
Vertex: (0, -3)

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Problem 10: y = x² - 2x - 3

Find zeros using formula: x = [-b ± √(b²-4ac)] / (2a)

a=1, b=-2, c=-3

Discriminant: b² - 4ac = (-2)² - 4(1)(-3) = 4 + 12 = 16

√16 = 4

x = [2 ± 4]/2

So:
x = (2+4)/2 = 6/2 = 3
x = (2-4)/2 = -2/2 = -1

Zeros: x = -1 and x = 3

Now plug back to find y (should be 0 since they’re zeros):

For x=-1: y = (-1)² -2(-1) -3 = 1 +2 -3 = 0 ✔
For x=3: y = 9 -6 -3 = 0 ✔

✔ Final for #10:
Zeros: x = -1, x = 3
Points: (-1, 0), (3, 0)

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Problem 11: y = 2x² + 6x

Factor out 2x: y = 2x(x + 3) → zeros when 2x=0 or x+3=0 → x=0 or x=-3

Or use formula: a=2, b=6, c=0

Discriminant: 36 - 0 = 36 → √36=6

x = [-6 ± 6]/(4)

x = (0)/4 = 0
x = (-12)/4 = -3

Same result.

Check y:
At x=0: y=0
At x=-3: y=2(9) +6(-3)=18-18=0

✔ Final for #11:
Zeros: x = 0, x = -3
Points: (0, 0), (-3, 0)

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Problem 12: y = 2x² + 6x ← Wait, same as #11?

Looking back at image: Problem 12 is written as “y = 2x² + 6x” — identical to #11.

Probably a copy-paste error in the worksheet. Maybe it was meant to be different? But as written, it’s the same.

So answer is same as #11.

✔ Final for #12:
Zeros: x = 0, x = -3
Points: (0, 0), (-3, 0)

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## Final Answers Summary:

4) Standard form: y = -3x²; Opens: down; Vertex: (0, 0); Axis: x = 0
5) Standard form: y = 9x² - x; Opens: up; Vertex: (1/18, -1/36); Axis: x = 1/18
6) Assuming typo: y = 3x² + 5x + 2; Opens: up; Vertex: (-5/6, -1/12); Axis: x = -5/6
7) Axis: x = -1; Vertex: (-1, -3)
8) Axis: x = 1; Vertex: (1, 4)
9) Axis: x = 0; Vertex: (0, -3)
10) Zeros: x = -1, 3; Points: (-1, 0), (3, 0)
11) Zeros: x = 0, -3; Points: (0, 0), (-3, 0)
12) Same as #11: Zeros: x = 0, -3; Points: (0, 0), (-3, 0)

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Final Answer:
See above summary for each problem's complete solution.