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

1 Answer
Aug 1, 2018

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

Explanation:

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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.