Graphing Quadratics Standard Form CW HW Rl4o07 | PDF - Free Printable

Let’s solve each problem step by step.

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Problem 1: f(x) = 3x²

This is a quadratic in standard form: f(x) = ax² + bx + c
Here, a = 3, b = 0, c = 0

✔ Vertex:
For any parabola f(x) = ax² + bx + c, the x-coordinate of the vertex is at x = -b/(2a)
→ x = -0/(2*3) = 0
Plug x=0 into f(x): f(0) = 3*(0)² = 0
So vertex is (0, 0)

✔ Axis of symmetry:
It’s the vertical line through the vertex → x = 0

✔ x-intercept(s):
Set f(x) = 0 → 3x² = 0 → x² = 0 → x = 0
Only one x-intercept: (0, 0)

✔ y-intercept:
Set x = 0 → f(0) = 0 → y-intercept is (0, 0)

✔ Max/Min:
Since a = 3 > 0, the parabola opens UP → so it has a MINIMUM

✔ Value of max/min:
The minimum value is the y-value of the vertex → 0

✔ Domain:
All real numbers → (-∞, ∞) or “all real numbers”

✔ Range:
Since minimum is 0 and it goes up forever → [0, ∞) or “y ≥ 0”

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Problem 2: f(x) = x² + 2x + 1

Again, standard form: a = 1, b = 2, c = 1

✔ Vertex:
x = -b/(2a) = -2/(2*1) = -1
f(-1) = (-1)² + 2*(-1) + 1 = 1 - 2 + 1 = 0
Vertex: (-1, 0)

✔ Axis of symmetry:
Vertical line through vertex → x = -1

✔ x-intercept(s):
Set f(x) = 0 → x² + 2x + 1 = 0
Factor: (x+1)(x+1) = 0 → x = -1 (double root)
Only one x-intercept: (-1, 0)

✔ y-intercept:
Set x = 0 → f(0) = 0 + 0 + 1 = 1 → (0, 1)

✔ Max/Min:
a = 1 > 0 → opens UP → MINIMUM

✔ Value of max/min:
Minimum value is y at vertex → 0

✔ Domain:
All real numbers → (-∞, ∞)

✔ Range:
Minimum is 0, goes up → [0, ∞) or “y ≥ 0”

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Final Answer:

Problem 1: f(x) = 3x²
- vertex: (0, 0)
- axis: x = 0
- x-int: (0, 0)
- y-int: (0, 0)
- max/min: min
- value: 0
- domain: all real numbers
- range: y ≥ 0

Problem 2: f(x) = x² + 2x + 1
- vertex: (-1, 0)
- axis: x = -1
- x-int: (-1, 0)
- y-int: (0, 1)
- max/min: min
- value: 0
- domain: all real numbers
- range: y ≥ 0