# How do you find the vertex and the intercepts for f(x)=x^2+7+6?

Nov 23, 2017

vertex ➝ (-3.5,-6.25)

x-ax is ➝ (-6,0) ∪ (-1,0)
y-ax is ➝ (0,6)

#### Explanation:

I guess you wanted to write $y = {x}^{2} + 7 x + 6$.

The vertex equation is:
${x}_{v} = \frac{- b}{2 a}$

substituting the values,
x_v=(-7)/(2·1)=-3.5

now if we substitute in the function equation,
f(-3.5)=(-3.5)^2+7·(-3.5)+6=-6.25

so the vertex is point $\left(- 3.5 , - 6.25\right)$.

We can get the intercept points on x-axis by equaling the function to $0$,
${x}^{2} + 7 x + 6 = 0$

x=(-b+-sqrt(b^2-4ac))/(2a)=(-7+-sqrt(7^2-4·1·6))/(2·1)

${x}_{1} = - 6$
${x}_{2} = - 1$

so point $\left(- 6 , 0\right)$ and $\left(- 1 , 0\right)$.

If we substitute $x$by $0$ we get the y-axis intercept point,
f(0)=0^2+7·0+6=6

point (0,6).