What is the vertex form of #y=-3x^2+ 17x + 2#?

1 Answer
Aug 1, 2018

#" "#
Vertex Form: #color(red)(y=f(x)=-3[x-17/6]^2+939/36#

Explanation:

#" "#
We are given the quadratic function in Standard Form:

#color(red)(y=f(x)=-3x^2+17x+2#

#color(blue)(y=f(x)=ax^2+bx+c#

What is expected?

We must convert to Vertex Form:

#color(blue)(y=f(x)=a(x-h)^2+k#

We have,

#y=f(x)=-3x^2+17x+2#

#color(green)("Step 1"#

Use Completing the Square Method to convert to Vertex Form:

#rArr (-3x^2+17x)+2#

#rArr -3[x^2-(17/3)x]+2#

#color(green)("Step 2"#

#rArr -3[x^2-(17/3)x+ square]+2#

In the #square# above, add #[(1/2)(17/3)]^2#

#rArr -3[x^2-(17/3)x+ [(1/2)(17/3)]^2]+2#

#rArr -3[x^2-(17/3)x+ (17/6)^2]+2#

#color(green)("Step 3"#

#rArr -3[x^2-(17/3)x+ (17/6)^2]+2- square#

Since we added #(17/6)^2# in the previous step, we must also subtract the same value.

#rArr -3[x^2-(17/3)x+ (17/6)^2]+2- (17/6)^2#

#color(green)("Step 4"#

On simplification, we get

#rArr -3[x^2-(17/3)x+ (17/6)^2]+(939/36)#

#y=f(x)= -3[x-17/6]^2+(939/36)#

Now, we have the required vertex form.

Hope this helps.