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f(x)=-(x+1)(x+7)
The function represents a parabola in the 
xy-plane. Which of the following is an equivalent form of 
f in which the 
y-intercept of the graph of 
f appears as a constant or coefficient?
Choose 1 answer:
(A) 
f(x)=-(x+4)^(2)+9
(B) 
f(x)=-(x+4)^(2)-9
(C) 
f(x)=-x^(2)+8x+7
(D) 
f(x)=-x^(2)-8x-7

f(x)=(x+1)(x+7)f(x) = -(x+1)(x+7)\newlineThe function represents a parabola in the xyxy-plane. Which of the following is an equivalent form of ff in which the yy-intercept of the graph of ff appears as a constant or coefficient?\newlineChoose 11 answer:\newline(A) f(x)=(x+4)2+9f(x) = -(x+4)^{2}+9\newline(B) f(x)=(x+4)29f(x) = -(x+4)^{2}-9\newline(C) f(x)=x2+8x+7f(x) = -x^{2}+8x+7\newline(D) f(x)=x28x7f(x) = -x^{2}-8x-7

Full solution

Q. f(x)=(x+1)(x+7)f(x) = -(x+1)(x+7)\newlineThe function represents a parabola in the xyxy-plane. Which of the following is an equivalent form of ff in which the yy-intercept of the graph of ff appears as a constant or coefficient?\newlineChoose 11 answer:\newline(A) f(x)=(x+4)2+9f(x) = -(x+4)^{2}+9\newline(B) f(x)=(x+4)29f(x) = -(x+4)^{2}-9\newline(C) f(x)=x2+8x+7f(x) = -x^{2}+8x+7\newline(D) f(x)=x28x7f(x) = -x^{2}-8x-7
  1. Expand and Simplify: To find the y-intercept of the graph of f, we need to express f(x)f(x) in a form where x=0x = 0 yields the y-intercept directly. This means we need to expand the given product and simplify.\newlineCalculation: \newlinef(x)=(x+1)(x+7)f(x) = -(x+1)(x+7)\newlinef(x)=(x2+7x+x+7)f(x) = -(x^2 + 7x + x + 7)\newlinef(x)=(x2+8x+7)f(x) = -(x^2 + 8x + 7)\newlinef(x)=x28x7f(x) = -x^2 - 8x - 7
  2. Check Options: Now, we check the options to see which one matches the expanded form of f(x)f(x) and has the yy-intercept as a constant or coefficient.\newlineOption (A) f(x)=(x+4)2+9f(x) = -(x+4)^2 + 9 does not match because when expanded it would not yield x28x7-x^2 - 8x - 7.\newlineOption (B) f(x)=(x+4)29f(x) = -(x+4)^2 - 9 does not match for the same reason as (A).\newlineOption (C) f(x)=x2+8x+7f(x) = -x^2 + 8x + 7 is incorrect because the signs of the xx terms are positive, not negative.\newlineOption (D) f(x)=x28x7f(x) = -x^2 - 8x - 7 matches our expanded form and has the yy-intercept (7)(-7) as a constant.

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