<PRE><B>Question 230804</B>
find the y-intercept of a parabola whose vertex is (-1,2), which is turned
 downward, and whose width is the same as y = 2x^2.
:
Using the form ax^2 - bx + c = y
:
We know the axis of symmetry; x = -1, and a=2
:
use the axis of symmetry formula to find b; x = -b/(2*a)
-1 = {{{(-b)/(2*2)}}}
-1 = {{{(-b)/4}}}
4(-1) = -b
-4 = -b
b = +4
:
We have 2x^2 + 4x + c = y
:
when x=-1, y=2
2(-1)^2 + 4(-1) + c = 2
2(1) - 4 + c = 2
2 - 4 + c = 2
-2 + c = 2
c = 2 + 2
c = 4
:
equation: y = 2x^2 + 4x + 4; therefore, the y intercept = 4
:
looks like this:
{{{ graph( 300, 200, -6, 5, -10, 10, 2x^2+4x+4)) }}} 
Vertex -1,2; y intercept +4
</PRE>