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Answer:
a) a=1, b=4, c=-5
b) opens up; a > 0
c) minimum = -9, the y-value of the vertex
d) x = -5, 1
e) y = -5, the value of c
f) (-2, -9)
Explanation:
A graph drawn by a graphing calculator can answer all of these questions quickly and easily.
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a) The values of 'a', 'b', and 'c' are the values of the coefficients of the terms in the function definition. They refer to the coefficients in the standard form:
f(x) = ax^2 +bx +c
f(x) = x^2 +4x -5 . . . . . . . your equation with a=1, b=4, c=-5
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b) The parabola opens up. This is because the value of 'a' is greater than zero.
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c) The minimum is the y-value of the vertex. When the function is written in vertex form it is ...
f(x) = (x -h)^2 +k . . . . . . . . . . . vertex (h, k)
Your function can be put into this form:
f(x) = x^2 + 4x +4 -5 -4 . . . . . . add and subtract 4 to complete the square
f(x) = (x +2)^2 -9 . . . . . . . . vertex form. The vertex is (-2, -9).
The y-value of the vertex is -9, the minimum value of the function.
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d) Solving f(x) = 0, you get ...
(x +2)^2 -9 = 0
(x +2)^2 = 9 . . . . . . . . add 9
x +2 = ±√9 = ±3 . . . . . take the square root; next, subtract 2
x = -2 ±3 = {-5, 1} . . . . . the x-intercepts
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e) The y-intercept is f(0) = c = -5.
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f) The vertex was found in part (c) to be (-2, -9).