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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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 =-(6)/2(1)
y = (x+1)(x+5), I guess you can solve the rest by yourself

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f(x)+x^2+6x+5 can be expressed as y = (x+1)(x+5)
Um okay.. This is my first problem with this. And I basically have no clue as to what im doing. And what your saying doesnt follow my examples and you haven't said anything about what ive already solved.
It says to solve for the vertex with x=-b/2a Then plug the answer for x into the problem to get the y-intercept. Factor, then set each factor to 0, that gives the answer x-intercept.
it's a point where you get the maxima or minima
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hope it helps!
I guess.. Now what?
intercepts (-5,0) and (-1,0) for vertex put x = -3 in the equation find y y = (-3+1)(-3+5) = -4 so vertex (-3,-4)
Um why are you doing it with y? My examples show to do it like this-- f(-3)= (-3)^2 -6(-3)+5
y = f(x) they are both the same
we write it like this >> y = f(x) = (x+1)(x+5) y is a function in x

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