OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):
5 years ago

OpenStudy (stamp):

You are given roots and a vertex.

5 years ago
OpenStudy (anonymous):

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?

5 years ago
OpenStudy (stamp):

roots at x = 3/2 and x = 5/2, these are the solutions (x - 3/2)(x - 5/2)=0 \[x^2-4x+15/4\]

5 years ago
OpenStudy (stamp):

sorry the other root is 9/2

5 years ago
OpenStudy (stamp):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (stamp):

(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

5 years ago
OpenStudy (stamp):

\[x^2-6x+27/4\] -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

5 years ago
OpenStudy (stamp):

\[x^2-6x+4\]\[f(3)=-5\]\[f(4)=16-24+4=-4\] yeah idk man how are you suposed to do this

5 years ago
OpenStudy (stamp):

is this a calculus class?

5 years ago
OpenStudy (anonymous):

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.

5 years ago
OpenStudy (stamp):

the y intercept looks like it's ~ 12

5 years ago
OpenStudy (anonymous):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (stamp):

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

5 years ago
OpenStudy (stamp):

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

5 years ago
OpenStudy (stamp):

y=a(x-3)-5

5 years ago
OpenStudy (stamp):

we find a using the other point

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (stamp):

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

5 years ago
OpenStudy (stamp):

imo

5 years ago
OpenStudy (stamp):

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

5 years ago
OpenStudy (anonymous):

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 :/

5 years ago
OpenStudy (stamp):

ok that is the answer tho

5 years ago
OpenStudy (stamp):
5 years ago

OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

Is this the right answer?

5 years ago
OpenStudy (stamp):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago