# What is the the vertex of y = (x -3)^2+4x-5 ?

Jan 6, 2016

The solution set(or vertex set) is: $S = \left\{- 5 , - 21\right\} .$

#### Explanation:

The standard formula of the quadratic function is:
$y = A {x}^{2} + B x + C$

${\left(x - 3\right)}^{2}$ is a notable product, so do this:
Square the first number -(signal inside the parenthesis) 2 * first number * second number + second number squared
${x}^{2} - 6 x + 9$

Now, substitute it the main equation:
$y = {x}^{2} - 6 x + 9 + 4 x - 5 = {x}^{2} + 10 x + 4$, so
$y = {x}^{2} + 10 x + 4$ $\to$ Now, it agrees with the standard formula.

To find the point of the vertex in $x$ axis, we apply this formula:
${x}_{v e r t e x} = - \frac{b}{2 a} = - \frac{10}{2} = - 5$

To find the point of the vertex in $y$ axis, we apply this formula:
${y}_{v e r t e x} = - \frac{\triangle}{4 a} = - \frac{{b}^{2} - 4 a c}{4 a} = - \frac{100 - 4 \cdot 1 \cdot 4}{4} = - 21$

Then, the solution set(or vertex set) is: $S = \left\{- 5 , - 21\right\} .$