+0

# parabola

0
55
1

What is the standard form of the quadratic function that has a vertex at (3,-2) and goes through the point (4,0)?

Apr 8, 2020

#1
+109520
+1

What is the standard form of the quadratic function that has a vertex at (3,-2) and goes through the point (4,0)?

Here is the easy way:

the axis of symmetry is x=3 with vertex at y=-2

one root is at 4

3+1=4

3-1=2

so the other root is at 2

\(y=k(x-2)(x-4)\\ sub\;in\; (3,-2)\\ -2=k*1*-1\\ k=2\\ y=2(x-2)(x-4)\\ y=2(x^2-6x+8)\\ y=2x^2-12x+16\)

Or if you prefer the harder way.  then...

\((x-h)^2=4a(y-k)\\ (x-3)^2=4a(y+2)\\ sub\;\; in\;\; (4,0)\\ (4-3)^2=4a(0+2)\\ 1=8a\\ a=0.125\\~\\ (x-3)^2=4*0.125(y+2)\\ x^2-6x+9=0.5y+1\\ x^2-6x+8=0.5y\\ y=2x^2-12x+16 \)

Coding:

y=k(x-2)(x-4)\\
sub\;in\; (3,-2)\\
-2=k*1*-1\\
k=2\\
y=2(x-2)(x-4)\\
y=2(x^2-6x+8)\\
y=2x^2-12x+16

(x-h)^2=4a(y-k)\\
(x-3)^2=4a(y+2)\\
sub\;\; in\;\; (4,0)\\
(4-3)^2=4a(0+2)\\
1=8a\\
a=0.125\\~\\
(x-3)^2=4*0.125(y+2)\\

x^2-6x+9=0.5y+1\\

x^2-6x+8=0.5y\\

y=2x^2-12x+16

Apr 8, 2020