Given the function f(x) = -5(x+4)^2 +15 complete the following. Please show all of your work. i. If f(x) = 15, what is x? ii. What point(s) from part (i) are on the graph of f? iii. List the x-intercepts, if any, of the graph of f. iv. List the y-intercept, if there is one, of the graph of f. Given the function f(x) = -5(x+4)^2 +15 complete the following. Please show all of your work. i. If f(x) = 15, what is x? ii. What point(s) from part (i) are on the graph of f? iii. List the x-intercepts, if any, of the graph of f. iv. List the y-intercept, if there is one, of the graph of f. @Mathematics

i. f(x) = -5(x+4)^2 +15 15 = -5(x+4)^2 +15 15-15 = -5(x+4)^2 +15-15 0 = -5(x+4)^2 -5(x+4)^2 = 0 (x+4)^2 = 0/(-5) (x+4)^2 = 0 x+4 = sqrt(0) x+4 = 0 x = -4

ii. Since f(x) = 15 and y is the same as f(x), this means that y = 15 But we know that when f(x) = 15, then x = -4 So when x = -4, y = 15 So the point (-4, 15) lies on the graph

iii. f(x) = -5(x+4)^2 +15 0 = -5(x+4)^2 +15 -15 = -5(x+4)^2 -15/(-5) = (x+4)^2 3 = (x+4)^2 (x+4)^2 = 3 x+4 = +-sqrt(3) x+4 = sqrt(3) or x+4 = -sqrt(3) x= -4 + sqrt(3) or x = -4 -sqrt(3) x = -2.2679 or x = -5.7321 So the x-intercepts are approximately (-2.2679 , 0) and (-5.7320, 0)

iv. f(x) = -5(x+4)^2 +15 f(0) = -5(0+4)^2 +15 f(0) = -5(4)^2 +15 f(0) = -5(16) +15 f(0) = -80 +15 f(0) = -65 So the y-intercept is (0, -65)

Wow! Awesome. Thanks!!

yw

Join our real-time social learning platform and learn together with your friends!