Question

Find the equation of the quadratic function f whose graph is shown below.

Answer 1

You are given roots and a vertex.

Answer 2

I don't know where to start. I have the root and the vertex, but do I use y=a(-h)^2+k to find the third point, or y=ax^2+bx=c?

Answer 3

roots at x = 3/2 and x = 5/2, these are the solutions

(x - 3/2)(x - 5/2)=0
\[x^2-4x+15/4\]

Answer 4

sorry the other root is 9/2

Answer 5

so 3/2 and 9/2 you can fix the factoring

Answer 6

So it would be x^2-4x +27/4 ??

Answer 7

(x-3/2)(x-9/2)
\[x^2-3/2\ x-9/2\ x+27/4\]\[x^2-12/2x+27/4\]

surely you can simplify that

Answer 8

\[x^2-6x+27/4\]
-b/2a is the x value of the vertex

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

the vertex is at 3

check the f(3)

f(3) = 9 - 18 + 27/4 = -9 + ~6 = -3

so idk the f(3) seems off maybe it is

Answer 9

\[x^2-6x+4\]\[f(3)=-5\]\[f(4)=16-24+4=-4\]

yeah idk man how are you suposed to do this

Answer 10

is this a calculus class?

Answer 11

lol- I'm looking stamp. This one is tough! It's an Aleks question on their assessment so it just keeps getting harder and harder.

Answer 12

the y intercept looks like it's ~ 12

Answer 13

It's going to take me a second... Maybe I can graph it using excel to find the missing formula, but sometimes that doesn't work either.

Answer 14

Thats what I thought. I tried 12, 0 as the y intercept but it threw the parabola off way right when I regraphed it....

Answer 15

y=a(x-h)^2 + k

Answer 16

(h,k) is the vertex 3, -5

Answer 17

y=a(x-3)-5

Answer 18

we find a using the other point

Answer 19

If i used that formula as the known vertex (3,-5) i still have to close the quadratic equation with a known third point. AAGGGGHHHH

Answer 20

\[y=a(x-3)^2-5\]\[a=(y+5)/(x-3)^2\]\[(x,y)=4,-3\]\[a=(-3+5)/(4-3)^2=2/1=2\]\[y=2(x-3)^2-5\]

Answer 21

imo

Answer 22

y-int (0, y) is (0, 2(0-3)^2-5)

is

(0, 2(9)-5)

is

(0, 13)

so the y int is 13

Answer 23

I'm going to have to press pause for a second Stamp. I have two sick little boys in the other room I have to tend to for a second :/

Answer 24

ok that is the answer tho

Answer 25

This is what I got as the equation:
\[y=2x ^{2}+12x+13 \]
So f(x)=\[2x ^{2}+12x+13 \]

Answer 26

Is this the right answer?

Answer 27

\[y=2(x-3)^2-5\]\[y=2(x^2-6x+9)-5\]\[y=(2x^2-12x+18)-5\]\[y=2x^2-12x+13\]

Answer 28

Thank you very much stamp! I sincerely appreciate it!!