How do you find the vertex and the intercepts for # f(x) = x² -10x -3#?

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Answer 1 · 2017-10-23T11:48:59

Vertex is at #(5,-28)#, x-intercepts are at
#(-0.29,0) and (10.29,0)# and y-intercept is at #(0,-3)#

#f(x)= x^2-10x-3 or f(x)= (x^2-10x+25)-25-3 # or

# f(x)=(x-5)^2-28 #. Comparing with vertex form of equation

#y=a(x-h)^2+k ; (h,k)# being vertex , #h=5 , k= -28 :.#

vertex is at #(5,-28)#, y-intercept is found by puuting

#x=0# in the equation # y= 0^2-10*0-3 = -3 :.#

y-intercept is #y=-3 or (0,-3)# . x-intercepts are found by

puuting #y=0# in the equation # 0=(x-5)^2-28 # or

# (x-5)^2=28 or x-5 = +- sqrt28 or x = 5 +- sqrt28 #

#:. x~~10.29 and x~~ -0.29 :. #x-intercepts are at

#(-0.29,0) and (10.29,0)#

graph{x^2-10x-3 [-80, 80, -40, 40]} [Ans]