How do you find the vertex and intercepts for #y=8(x-10)^2-16#?

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Answer 1 · 2017-09-05T16:51:46

The vertex form of a quadratic function is given by
f (x) = a(x - h)2 + k, where (h, k) is the vertex of the parabola.

So here, it's already in vertex form. a = 8, h = 10, k = -16, and the vertex coordinates are 10, -16

Intercepts can be easily calculated. The y intercept is found when x = 0 is substituted into the original equation:

#y = 8(-10)^2 - 16 = 800 - 16 = 784#

x intercepts are found by setting y = 0 and solving for x.

#0 = 8(x - 10)^2 - 16#

...easiest way to solve this is to convert it to a standard format quadratic equation.

Multiply out:

#8(x^2 - 20x + 100) - 16 = 0#

#8x^2 - 160x + 784 = 0#

...this is standard form, #ax^2 + bx + c = 0#, with
a = 8, b = -160, c = 784

giving roots 11.41 and 8.59 (rounded)

...always good to have a graph as function as a sanity check:

graph{8(x - 10)^2 - 16 [-10.51, 21.52, -19.47, -3.45]}

GOOD LUCK!

Answer 2 · 2017-09-05T16:55:06

Vertex is #(10,-16)#, #x#-intercept are at #(10-sqrt2,0)# and #(10+sqrt2,0)# and #y#-intercept is at #(0,784)#.

To find the vertex, we should have equation into vertex form i.e. #y=a(x-h)^2+k#, in which case vertex is #(h,k)#.

As #y=8(x-10)^2-16# is alreaddy in this form, with #a=8#, #h=10# and #k=-16#

Hence, vertex is #(10,-16)#

For #x#-intercepts, we should put #y=0# and for #y#-intercepts, we should put #x=0#

Putting #x=0#, #y=8(0-10)^2-16=784# and #y#-intercept is at #(0,784)#.

and when #y=0#, we have #8(x-10)^2-16=0# i.e. #(x-10)^2=2# i.e. #x=10+-sqrt2# and #x#-intercept are at #(10-sqrt2,0)# and #(10+sqrt2,0)#.

#color(red)("Graph not drawn to scale")#
graph{8(x-10)^2-16 [-5, 20, -100, 860]}