# What is the vertex form of y=5x^2-30x+49?

Jul 25, 2017

See a solution process below:

#### Explanation:

To convert a quadratic from $y = a {x}^{2} + b x + c$ form to vertex form, $y = a {\left(x - \textcolor{red}{h}\right)}^{2} + \textcolor{b l u e}{k}$, you use the process of completing the square.

First, we must isolate the $x$ terms:

$y - \textcolor{red}{49} = 5 {x}^{2} - 30 x + 49 - \textcolor{red}{49}$

$y - 49 = 5 {x}^{2} - 30 x$

We need a leading coefficient of $1$ for completing the square, so factor out the current leading coefficient of 2.

$y - 49 = 5 \left({x}^{2} - 6 x\right)$

Next, we need to add the correct number to both sides of the equation to create a perfect square. However, because the number will be placed inside the parenthesis on the right side we must factor it by $2$ on the left side of the equation. This is the coefficient we factored out in the previous step.

y - 49 + (5 * ?) = 5(x^2 - 6x + ?) <- Hint: $\frac{6}{2} = 3$; $3 \cdot 3 = 9$

$y - 49 + \left(5 \cdot 9\right) = 5 \left({x}^{2} - 6 x + 9\right)$

$y - 49 + 45 = 5 \left({x}^{2} - 6 x + 9\right)$

$y - 4 = 5 \left({x}^{2} - 6 x + 9\right)$

Then, we need to create the square on the right hand side of the equation:

$y - 4 = 5 {\left(x - 3\right)}^{2}$

Now, isolate the $y$ term:

$y - 4 + \textcolor{b l u e}{4} = 5 {\left(x - 3\right)}^{2} + \textcolor{b l u e}{4}$

$y - 0 = 5 {\left(x - 3\right)}^{2} + \textcolor{b l u e}{4}$

$y - 0 = 5 {\left(x - \textcolor{red}{3}\right)}^{2} + \textcolor{b l u e}{4}$

The vertex is: $\left(3 , 4\right)$

Jul 25, 2017

$y = 5 \left(x - 3\right) + 4$

#### Explanation:

$y = 5 {x}^{2} - 30 x + 49$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{30}{10} = 3$
y-coordinate of vertex:
$y \left(3\right) = 5 \left(9\right) - 30 \left(3\right) + 49 = 4$
Vertex (3, 4)
Vertex form of y:
$y = 5 {\left(x - 3\right)}^{2} + 4$