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

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Answer 1 · 2017-11-23T18:25:15

#vertex ➝ (-3.5,-6.25)#

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

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

point (0,6).