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

Oct 8, 2017

$\left(1 , 12\right) , x = - 1 , x = 3$

#### Explanation:

$\text{to find the intercepts, that is where the graph crosses}$
$\text{the x and y axes}$

• " let x = 0, in the equation for y-intercept"

• " let y = 0, in the equation for x-intercepts"

$x = 0 \to y = - 3 \left(- 3\right) \left(1\right) = 9 \leftarrow \textcolor{red}{\text{ y-intercept}}$

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

$\text{equate each factor to zero and solve for x}$

$x - 3 = 0 \Rightarrow x = 3$

$x + 1 = 0 \Rightarrow x = - 1$

$\Rightarrow x = - 1 , x = 3 \leftarrow \textcolor{red}{\text{ x-intercepts}}$

$\textcolor{b l u e}{\text{the axis of symmetry"" is at the midpoint of the}}$
$\text{x-intercepts}$

$\Rightarrow x = \frac{- 1 + 3}{2} = 1 \Rightarrow x = 1 \text{ is the axis of symmetry}$

$\text{the vertex lies on the axis of symmetry}$

$\Rightarrow \text{ x-coordinate of vertex is } x = 1$

$\text{substitute this value into the equation for y-coordinate}$

$\Rightarrow {y}_{\textcolor{red}{\text{vertex}}} = - 3 \left(- 2\right) \left(2\right) = 12$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(1 , 12\right)$
graph{-3(x-3)(x+1) [-40, 40, -20, 20]}

Oct 8, 2017

Vertex is at $\left(1 , 12\right)$, x intercepts are at $\left(- 1 , 0\right) \mathmr{and} \left(3 , 0\right)$ and
y intercept is at
$\left(0 , 9\right)$

#### Explanation:

$y = - 3 \left(x - 3\right) \left(x + 1\right) \mathmr{and} y = - 3 \left({x}^{2} - 2 x - 3\right)$ or

$y = - 3 \left({x}^{2} - 2 x\right) + 9 \mathmr{and} y = - 3 \left({x}^{2} - 2 x + 1\right) + 3 + 9$ or

$y = - 3 {\left(x - 1\right)}^{2} + 12$. Comparing with vertex form of equation

y=a(x-h)^2+k ; (h,k) being vertex we find here

$h = 1 , k = 12 \therefore$ Vertex is at $\left(1 , 12\right)$

x intercepts can be found by putting $y = 0$ in the equation

$\therefore y = - 3 \left(x - 3\right) \left(x + 1\right) \mathmr{and} - 3 \left(x - 3\right) \left(x + 1\right) = 0$ or

$\left(x - 3\right) \left(x + 1\right) = 0 \therefore \left(x - 3\right) = 0 \mathmr{and} x = 3 , \left(x + 1\right) = 0$ or

$x = - 1 \therefore$ x intercepts are at $\left(- 1 , 0\right) \mathmr{and} \left(3 , 0\right)$

y intercept can be found by putting $x = 0$ in the equation

$\therefore y = - 3 \left(x - 3\right) \left(x + 1\right) \mathmr{and} y = - 3 \left(0 - 3\right) \left(0 + 1\right) = 9 \therefore$

y intercept is at $\left(0 , 9\right)$

graph{-3(x-3)(x+1) [-40, 40, -20, 20]} [Ans]