Question

can someone help me find the quadratic equation?
1. maximum value 9; zeros of f are -6 and 0.
i only know how to do it when given the x intercepts, not zero. thanks!

Answer 1

\(\large\color{black}{ \displaystyle y=a\cdot (x-h)^2+k }\)

that is a form of an opening down parabola,
(where a is a real number, such that, a<0)
with vertex (or maximum in our case) of (h,k).

Answer 2

So far you know that k=9 (that is your maximum).
\(\large\color{black}{ \displaystyle y=(-a)\cdot (x-h)^2+9 }\)

and you know that (0,0) and (0,-6) are x-intercept

Answer 3

So, since parabola is symmetric you can right of
the bet conclude that it is centered at:

Center x-value =(-6+0)/2 = -3
(at x=-3)

So,

\(\large\color{black}{ \displaystyle y=(-a)\cdot (x--3)^2+9 }\)
\(\large\color{black}{ \displaystyle y=(-a)\cdot (x+3)^2+9 }\)

Answer 4

oh, I wrote -a, so a should be greater than zero,
for the coefficient -a to be neg.

Answer 5

Use any of the points to find a.

\(\large\color{black}{ \displaystyle 0=(-a)\cdot (0+3)^2+9 }\)
and solve for a.

Answer 6

ohhh ok. so whenever i am given zeros, i plug it into the first equation?

Answer 7

a=1?

Answer 8

Well, if you are given the zeros,
you know the vertex x value.

Because the parabola is always symmetric,
so if it has zeros of x=c and x=d, then the
x-value of the vertex is =(c+d)/2
or the midpoint X between the x-intercepts.

Answer 9

-9=-9a
a=1

precisely right!

Answer 10

also, do i always have to set a to -a?

Answer 11

Well, you can either say that it is -a, and solve for "a",
but when you are writing you would write a negative a coefficient.

\(\large\color{black}{ \displaystyle y=(-a)\cdot (x+3)^2+9 }\)
you found a=1, so
\(\large\color{black}{ \displaystyle y=(-\color{red}{1})\cdot (x+3)^2+9 }\)

Answer 12

OR////

Answer 13

Well, you can either say that it is a
\(\large\color{black}{ \displaystyle y=(a)\cdot (x+3)^2+9 }\)

then you would find that a=-1,
and you would put it right in there
\(\large\color{black}{ \displaystyle y=(\color{red}{-1})\cdot (x+3)^2+9 }\)

Answer 14

You still have that -1 coefficient...

Answer 15

Oh i see. THANK YOU SOO MUCH! You've been very helpful!!!

Answer 16

Anytime...
also do you know how I knew that my a is negative?

Answer 17

I am saying, do you know how I figured that the coefficient in front of x^2 must be negative?

Answer 18

Does it depend on whether the equation is minimum or maximum?

Answer 19

Yes, basically very good.
(Well if we ae taking about a vertically (up/down) opening parabola)

So if it is going down, the coefficient is negative
and if it is going up, the coefficient is positive

Answer 20

So an opening down parab
will have a maximum (but not min)

and vice versa.

opening up parab
will have a minimum (but not max)

Answer 21

I understand now. Thanks again!

Answer 22

You are welcome!