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

Question

Isaiah Anthony · Accepted Answer

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

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

Eva Abramson · Answer

undefined The important points needed to graph $ y = 8(x - 3)^2 - 5 $ are the vertex, which is (3, -5), and the direction of opening, which is upwards because the coefficient of the quadratic term is positive. Additionally, you can identify the y-intercept, which is (-3, 67).