How do you find the vertex and the intercepts for #y = -5x^2 + 10x - 7#?

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Answer 1 · 2017-02-09T05:57:41

#V (1:-2)#

one intercept #(0;-7)#

You can find the vertex by applying the following formula:

#V_x=-b/(2a)#

and then substitute its value in the equation to find #V_y#:

#V_x=-10/(2(-5))=1

#V_y=-5*1^2+10*1-7=-2#

Then the vertex is #V (1:-2)#

To find the intercepts with y-axis, you would substitute #x=0# in the equation and, with x-axis, #y=0#:

#x=0 -> y=-7#

#y=0->-5x^2+10x-7=0->x=(-5+-sqrt(25-35))/-5->#there are no real solutions
graph{-5x^2+10x-7 [-2,3, -10, 1]}