Respuesta :
[tex] f(x)=2x^2+32x+126 [/tex]
Graph the parabola by first plotting its vertex and then plotting a second point on the parabola.
To find x coordinate of vertex we use formula [tex] x =\frac{-b}{2a} [/tex]
From the given equation , a= 2 , b= 32, c= 126
Plug in all the values
[tex] x =\frac{-32}{2*2} [/tex] = -8
Now plug in 8 for x in f(x)
[tex] f(x)=2x^2+32x+126 [/tex]
[tex] f(-8)=2(-8)^2+32(-8)+126 [/tex] = -2
So vertex is (-8,-2)
Now we make a table to get some other points for graphing. We assume x value before and after vertex. Plug in -9 and -7 for x in f(x) and find out y .
[tex] f(x)=2x^2+32x+126 [/tex]
[tex] f(-9)=2(-9)^2+32(-9)+126 [/tex] = 0
[tex] f(-7)=2(-7)^2+32(-7)+126 [/tex] = 0
x ------> y
-9 ------> 0
-8 -------> -2
-7 --------> 0
Now we graph all the points.
Graph is attached below.

Here are a bunch of CORRECT answers, your answer is somewhere in there. For the first CORRECT answer the second point is -5,-9. Don't make the same mistake I did on question 3.




