Question

Can someone explain how to solve this? Thank you. Find the coordinates of the vertex and the intercepts of the following quadratic function. When necessary, approximate the x-intercepts to the nearest tenth.

Answer 1

f(x)+x^2+6x+5

Answer 2

a= 1 b= 6 c= 5

x= -(6)/2(1)
=-6/2
=-3

Answer 3

@pooja195

Answer 4

that's the x coordinate of the vertex.

Answer 5

to find the y coordinate you need to put x =-3 in the equation and solve for the value of y.

Answer 6

f(-3)=(-3)^2 -6(-3)+5
= 9+18+5
32

Answer 7

where y is nothing but f(x) itself.
and find the ordered pair as V(x,f(x))

Answer 8

f(x)=x^2+6x+5
f(-3) = (-3)^2 +6*(-3) +5 = 9-18+5 =4

Answer 9

vertex= (-3,-4) ?

Answer 10

yeah that will be -4 not 4 great, yeah! that's your vertex.

Answer 11

Okay...
For y-intercept

f(0)=(0)^2+6(0)+5
5

(0,5) ?

Answer 12

yes!

Answer 13

Okay

Answer 14

for x intercepts you need to put x^2+6x+5 = 0 and find all possible x's.

Answer 15

Okay so I have to factor x^2+6x+5=0

Answer 16

yes!
(x+1)(x+5) =0; x=-1,-5 and f(x)'s at these x's are 0.

Answer 17

Okay.. Is that everything?

Answer 18

For this question.

Answer 19

Okay... Thank you.

Answer 20

:) wlcm