Horizontal stretch of 2. 2 Units up. Reflection in y-axis. Vertex

Question

The graph of  is a horizontal stretch by a factor of 2 and a translation 2 units up, followed by a reflection in the y-axis of the graph of f(x)=8x^2−6. Write a rule for g. Then identify the vertex.

Answer 1

^{The tutor that tried to answer this question already did a great job at describing what all the different translations do, so there is no need to elucidate upon that aspect of this problem. It seems that he came up with an incorrect answer because there was a miscommunication with what the original function is.}

given that f(x) = 8x² - 6

g(x) = 8[-1/2(x)²] - 6 + 2

g(x) = -4x² - 4

for the vertex (h,k)

h = 0/-8 = 0

k= g(h) = g(0) = -4

(h,k) = (0,-4)

Answer 2

Horizontal Stretches/Compressions - multiply the x value directly. For a horizontal stretch of 2, x² would become (x/2)². Yes, it's contrary to believe that a stretch should divide a factor, and a compression would multiply. But do not divide outside of the parenthesis, it remains close to the X.

Vertical Translations - add or subtract a factor to the end of the equation. f(x) translated up 2 would be f(x)+2

Reflection over Y-axis would mean that all X-values will switch their sign. Reflecting x² would be (-x)². Any changes that affects x directly must be "within parenthesis," or close to X. If this was done over the x-axis, the resulting graph would've been -(x)², where the negative sign is outside.

For this equation:

f(x) = 8x² + 6

g(x) = 8(-x/2)²+ 8 -> g(x) = 2x² + 8

Vertex formula of x-coordinate: (-b/2a)

b = 0

a = 8

(0/16) = 0

Plug in x =0 to find y coordinate

vertex = (0,8)