What are the important points needed to graph #y = 8(x - 3)^2 - 5#?

1 Answer
Jul 3, 2018

vertex #(3, -5)#
#x#-intercepts: #(3 - sqrt(10)/4, 0), (3 + sqrt(10)/4, 0)#

#y#-intercept: (#0, 67)#

Explanation:

Given: #y = 8(x-3)^2 - 5#

#color(blue)("Find the vertex:")#

From the vertex form: #y = a(x - h)^2 + k#, where vertex #(h, k)#

we can find the vertex of the given equation as #(3, -5)#

#color(blue)("Find the "y-"intercept")# by setting #x = 0#:

#y = 8(0-3)^2 - 5#

#y = 8*9 - 5 = 72 - 5 = 67#

#y#-intercept: (#0, 67)#

#color(blue)("Find the "x-"intercepts")# by setting #y = 0#:

#0 = 8(x-3)^2 - 5#

Distribute using the square function: #(a + b)^2 = a^2 + 2ab + b^2#

#8 (x^2 - 6x + 9) - 5 = 0#

#8x^2 - 48x + 72 - 5 = 0#

#8x^2 - 48x + 67 = 0#

Use the quadratic formula to solve:

#x =( -B +- sqrt(B^2 - 4AC))/(2A)#,

where the equation is in the form: #Ax^2 + Bx + C = 0#

#x =( 48 +- sqrt((-48)^2 - 4*8*67))/(16)#

#x = 48/16 +- sqrt(160)/16#

#x = 3 +- (sqrt(16 * 10))/16#

#x = 3 +- (4 sqrt(10))/16#

#x = 3 +- sqrt(10)/4#

#x#-intercepts: #(3 - sqrt(10)/4, 0), (3 + sqrt(10)/4, 0)#