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1)Graph the quadratic function 𝑓(𝑥) = (𝑥 + 2)2 − 1 a) Determine if parabola opens upward or downward b) Determine the vertex of the parabola c) find any x intercepts if exist by solving f(x)=0 d) find the y intercept by computing f(0) e) plot the x and y intercepts, vertex and additional point as necessary so connect point and graph the quadratic function above tips To graph 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘, 𝑎 ≠ 0 1. Determine whether the parabola opens upward or downward. If a > 0, it opens upward. If a < 0, it opens downward. 2. Determine the vertex of the parabola. The vertex is (h, k). 3. Find any x-intercepts by solving f(x) = 0. The function’s real zeros are the x-intercepts. 4. Find the y-intercept by computing f(0). 5. Plot the intercepts, the vertex, and additional points as necessary. Connect these points with a smooth curve that is shaped like a bowl or an inverted bowl.

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Answer to a math question 1)Graph the quadratic function 𝑓(𝑥) = (𝑥 + 2)2 − 1 a) Determine if parabola opens upward or downward b) Determine the vertex of the parabola c) find any x intercepts if exist by solving f(x)=0 d) find the y intercept by computing f(0) e) plot the x and y intercepts, vertex and additional point as necessary so connect point and graph the quadratic function above tips To graph 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘, 𝑎 ≠ 0 1. Determine whether the parabola opens upward or downward. If a > 0, it opens upward. If a < 0, it opens downward. 2. Determine the vertex of the parabola. The vertex is (h, k). 3. Find any x-intercepts by solving f(x) = 0. The function’s real zeros are the x-intercepts. 4. Find the y-intercept by computing f(0). 5. Plot the intercepts, the vertex, and additional points as necessary. Connect these points with a smooth curve that is shaped like a bowl or an inverted bowl.

Expert avatar
Jon
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109 Answers
a) Since a = 1 > 0, the parabola opens upward.

b) The vertex form of a quadratic is f(x) = a(x - h)^2 + k. Here, h = -2 and k = -1, hence the vertex is (-2, -1).

c) To find x-intercepts, solve f(x) = 0:

(x + 2)^2 - 1 = 0
(x + 2)^2 = 1

Taking square root: x + 2 = \pm 1

Solve this gives:
x + 2 = 1 \Rightarrow x = -1
x + 2 = -1 \Rightarrow x = -3

Thus, the x-intercepts are (-1, 0) and (-3, 0).

d) The y-intercept is found by setting x = 0:

f(0) = (0 + 2)^2 - 1 = 4 - 1 = 3

So the y-intercept is (0, 3).

e) Plotting points and drawing a smooth curve:

Vertex: (-2, -1)
X-intercepts: (-1, 0) \, \text{and} \, (-3, 0)
Y-intercept: (0, 3)

Connect these points with a smooth curve that is shaped like a bowl.

**Answer:**

\text{Vertex: } (-2, -1)

\text{X-intercepts: } (-1, 0), \, (-3, 0)

\text{Y-intercept: } (0, 3)

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