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

1 Answer
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+7x+6.

The vertex equation is:
x_v=(-b)/(2a)

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 (-3.5,-6.25).

We can get the intercept points on x-axis by equaling the function to 0,
x^2+7x+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 (-6,0) and (-1,0).

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

point (0,6).