What is the standard form of the quadratic function that has a vertex at (3,-2) and goes through the point (4,0)?
+118703 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)subin(3,−2)−2=k∗1∗−1k=2y=2(x−2)(x−4)y=2(x2−6x+8)y=2x2−12x+16
Or if you prefer the harder way. then...
(x−h)2=4a(y−k)(x−3)2=4a(y+2)subin(4,0)(4−3)2=4a(0+2)1=8aa=0.125 (x−3)2=4∗0.125(y+2)x2−6x+9=0.5y+1x2−6x+8=0.5yy=2x2−12x+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
