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

Aug 1, 2018

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

#### Explanation:

$\text{ }$
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,

$y = f \left(x\right) = - 3 {x}^{2} + 17 x + 2$

color(green)("Step 1"

Use Completing the Square Method to convert to Vertex Form:

$\Rightarrow \left(- 3 {x}^{2} + 17 x\right) + 2$

$\Rightarrow - 3 \left[{x}^{2} - \left(\frac{17}{3}\right) x\right] + 2$

color(green)("Step 2"

$\Rightarrow - 3 \left[{x}^{2} - \left(\frac{17}{3}\right) x + \square\right] + 2$

In the $\square$ above, add ${\left[\left(\frac{1}{2}\right) \left(\frac{17}{3}\right)\right]}^{2}$

$\Rightarrow - 3 \left[{x}^{2} - \left(\frac{17}{3}\right) x + {\left[\left(\frac{1}{2}\right) \left(\frac{17}{3}\right)\right]}^{2}\right] + 2$

$\Rightarrow - 3 \left[{x}^{2} - \left(\frac{17}{3}\right) x + {\left(\frac{17}{6}\right)}^{2}\right] + 2$

color(green)("Step 3"

$\Rightarrow - 3 \left[{x}^{2} - \left(\frac{17}{3}\right) x + {\left(\frac{17}{6}\right)}^{2}\right] + 2 - \square$

Since we added ${\left(\frac{17}{6}\right)}^{2}$ in the previous step, we must also subtract the same value.

$\Rightarrow - 3 \left[{x}^{2} - \left(\frac{17}{3}\right) x + {\left(\frac{17}{6}\right)}^{2}\right] + 2 - {\left(\frac{17}{6}\right)}^{2}$

color(green)("Step 4"

On simplification, we get

$\Rightarrow - 3 \left[{x}^{2} - \left(\frac{17}{3}\right) x + {\left(\frac{17}{6}\right)}^{2}\right] + \left(\frac{939}{36}\right)$

$y = f \left(x\right) = - 3 {\left[x - \frac{17}{6}\right]}^{2} + \left(\frac{939}{36}\right)$

Now, we have the required vertex form.

Hope this helps.