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.

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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.

Mathematics
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f(x)+x^2+6x+5
a= 1 b= 6 c= 5 x= -(6)/2(1) =-6/2 =-3

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Other answers:

that's the x coordinate of the vertex.
to find the y coordinate you need to put x =-3 in the equation and solve for the value of y.
f(-3)=(-3)^2 -6(-3)+5 = 9+18+5 32
where y is nothing but f(x) itself. and find the ordered pair as V(x,f(x))
f(x)=x^2+6x+5 f(-3) = (-3)^2 +6*(-3) +5 = 9-18+5 =4
vertex= (-3,-4) ?
yeah that will be -4 not 4 great, yeah! that's your vertex.
Okay... For y-intercept f(0)=(0)^2+6(0)+5 5 (0,5) ?
yes!
Okay
for x intercepts you need to put x^2+6x+5 = 0 and find all possible x's.
Okay so I have to factor x^2+6x+5=0
yes! (x+1)(x+5) =0; x=-1,-5 and f(x)'s at these x's are 0.
Okay.. Is that everything?
For this question.
Okay... Thank you.
:) wlcm

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