How do you find the vertex and intercepts for #y = 10x – 3x^2#?

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Answer 1 · 2016-06-17T16:28:52

The vertex is #(5/3,25/3)#
The intercepts are (0,0) and #(10/3,0)#

you can find the coordinate x of the vertex using the formula
#x=-b/(2a)#
where a is the coefficient of #x^2# and b of x

Since a=-3 and b=10, you have

#x=-10/-6=5/3#

Then you can put this value in x and find y
#y=10(5/3)-3(5/3)^2#

#y=50/3-3(25/9)#
#y=50/3-25/3#
#y=25/3#

One of the intercepts is the origin (0,0) because y has no known term

By solving the equation #10x-3x^2=0# you can find the other one:

#x(10-3x)=0#

#10-3x=0#

#x=10/3#