How do you find the vertex and the intercepts for $y = x^{2} + 6 x + 5$ $y=x^2+6x+5$ ?

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Answer 1 · 2017-08-07T14:42:50

Here is what I would do:

$y$ $y$ intecept

When $x = 0$ $x = 0$ , we get $y = 5$ $y = 5$

Vertex and $x$ $x$ Intercepts

Note that the following method will not work if the equation has no $x$ $x$ intercepts.

Also note that there are other ways to find the vertex.

$x$ $x$ Intercepts

x^2+6x+5 = 0#

$(x + 1) (x + 5) = 0$ $(x+1)(x+5) = 0$

$(x + 1) = 0$ $(x+1) = 0$ $" "$ OR $" "$ $(x + 5) = 0$ $(x+5) =0$

$x = - 1$ $x=-1$ or $- 5$ $-5$

the $x$ $x$ intercepts are $- 1$ $-1$ and $- 5$ $-5$ .

Vertex

The vertex is midway between the $x$ $x$ intercepts. So it is at the average of the $x$ $x$ intercepts.

The average is $\frac{1}{2}$ $1/2$ of the sum, so the vertex is at

$x = \frac{1}{2} ((1 -) + (- 5)) = \frac{- 6}{2} = - 3$ $x = 1/2((1-)+(-5)) = (-6)/2 = -3$

To find the $y$ $y$ coordinate of the vertex, plug $- 3$ $-3$ in for $x$ $x$ in the equation

$y = x^{2} + 6 x + 5$ $y = x^2+6x+5$ at $x = - 3$ $x = -3$ is

${(- 3)}^{2} + 6 (- 3) + 5 = 9 - 18 + 5 = - 9 + 5 = - 4$ $(-3)^2+6(-3)+5 = 9-18+5 = -9+5=-4$

The vertex is $(- 3, - 4)$ $(-3,-4)$

How do you find the vertex and the intercepts for y=x2+6x+5y=x^2+6x+5?