# How do you find the vertex and the intercepts for y= -3(x-1)(x+5)?

Vertex is at $\left(- 2 , 27\right)$
Intercepts at $\left(1 , 0\right)$, $\left(- 5 , 0\right)$,$\left(0 , 15\right)$

#### Explanation:

From the given $y = - 3 \left(x - 1\right) \left(x + 5\right)$
Expand the right side so that it will take the form $y = a {x}^{2} + b x + c$

$y = - 3 \left(x - 1\right) \left(x + 5\right)$

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

$y = - 3 {x}^{2} - 12 x + 15$

Now it takes the form $y = a {x}^{2} + b x + c$

with $a = - 3$ and $b = - 12$ and $c = 15$

We can now solve for the vertex $\left(h , k\right)$ using the formula

$h = - \frac{b}{2 a}$ and $k = c - {b}^{2} / \left(4 a\right)$

$h = \frac{- \left(- 12\right)}{2 \left(- 3\right)} = \frac{12}{-} 6 = - 2$

$k = c - {b}^{2} / \left(4 a\right) = 15 - {\left(- 12\right)}^{2} / \left(4 \left(- 3\right)\right) = 15 + \frac{144}{12} = 15 + 12 = 27$

Vertex $\left(h , k\right) = \left(- 2 , 27\right)$

To solve for the intercept, let us use the given $y = - 3 \left(x - 1\right) \left(x + 5\right)$

Solve for x-intercept by setting $y = 0$

$y = - 3 \left(x - 1\right) \left(x + 5\right)$

$0 = - 3 \left(x - 1\right) \left(x + 5\right)$

$0 = \left(x - 1\right) \left(x + 5\right)$

Equate both factors to 0
First factor
$\left(x - 1\right) = 0$
$x = 1$ there is an intercept at $\left(1 , 0\right)$

Second factor
$x + 5 = 0$
$x = - 5$ there is an intercept at $\left(- 5 , 0\right)$

Use again the original equation $y = - 3 \left(x - 1\right) \left(x + 5\right)$ to solve for y-intercept and setting $x = 0$

$y = - 3 \left(x - 1\right) \left(x + 5\right)$

$y = - 3 \left(0 - 1\right) \left(0 + 5\right)$

$y = - 3 \left(- 1\right) \left(5\right)$

$y = 15$

there is an intercept at $\left(0 , 15\right)$

The graph of $y = - 3 \left(x - 1\right) \left(x + 5\right)$ where you can see the
vertex -2, 27), and intercepts at $\left(1 , 0\right)$, $\left(- 5 , 0\right)$,$\left(0 , 15\right)$

graph{y=-3(x-1)(x+5)[-60,60,-30,30]}

God bless .... I hope the explanation is useful.