Asked by LOL
Match each quadratic function to its graph.
f(x)=x2+2
g(x)=2x2–16x+30=2(x–3)(x–5)
h(x)=
–
x2+2x–1=
–
(x–1)2
k(x)=2x2+16x+32=2(x+4)2
All Answers 1
Answered by
GPT-5 mini
AI
Give each parabola these identifying features so you can match to the pictures:
f(x) = x^2 + 2
- vertex (0, 2) (axis x = 0), opens upward
- y-intercept = 2
- no real x-intercepts (does not cross the x-axis)
g(x) = 2x^2 – 16x + 30 = 2(x–3)(x–5)
- opens upward
- x-intercepts at x = 3 and x = 5
- vertex at x = 4 with value g(4) = −2 → vertex (4, −2)
- y-intercept = 30
h(x) = −x^2 + 2x − 1 = −(x − 1)^2
- opens downward (maximum)
- double root at x = 1 (touches x-axis at (1,0))
- vertex (1, 0)
- y-intercept = −1
k(x) = 2x^2 + 16x + 32 = 2(x + 4)^2
- opens upward
- double root at x = −4 (touches x-axis at (−4,0))
- vertex (−4, 0)
- y-intercept = 32
Match each graph by checking (1) whether it opens up or down, (2) the vertex coordinates, and (3) whether it crosses/touches the x-axis and where. If you paste or describe the four graphs (their vertices or intercepts), I’ll tell you the exact pairing.
f(x) = x^2 + 2
- vertex (0, 2) (axis x = 0), opens upward
- y-intercept = 2
- no real x-intercepts (does not cross the x-axis)
g(x) = 2x^2 – 16x + 30 = 2(x–3)(x–5)
- opens upward
- x-intercepts at x = 3 and x = 5
- vertex at x = 4 with value g(4) = −2 → vertex (4, −2)
- y-intercept = 30
h(x) = −x^2 + 2x − 1 = −(x − 1)^2
- opens downward (maximum)
- double root at x = 1 (touches x-axis at (1,0))
- vertex (1, 0)
- y-intercept = −1
k(x) = 2x^2 + 16x + 32 = 2(x + 4)^2
- opens upward
- double root at x = −4 (touches x-axis at (−4,0))
- vertex (−4, 0)
- y-intercept = 32
Match each graph by checking (1) whether it opens up or down, (2) the vertex coordinates, and (3) whether it crosses/touches the x-axis and where. If you paste or describe the four graphs (their vertices or intercepts), I’ll tell you the exact pairing.
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