have a Graph with the vertex as a 'minimum' and with a turning point of 'x' intercept 5 and -2 and a 'y' intercept of -10. Turning point is -12y & 1.5x.
I need to find 'equations' that fit this graph and give two reasons for my answer.
I'm very confused with this one can I PLEASE get some help. I wish I could show you the graph as it will make it much easier.
Thank you!
To have x-intercepts at 5 and -2, it must be such that if you let
y=0 and solve for x you get
x = 5; and x = -2
That would be gotten be setting
x-5 = 0; and x+2 = 0
That means you must have had to use the zero factor principle on
(x-5)(x+2)=0
And there could have been any constant times that
and we would have gotten the same two x-intercepts.
So the equation must have been
y = a(x-5)(x+2)
y = a(x²-3x-10)
ince it has y-intercept -10, when we set x=0, y equals -10
-10 = a(0²-3·0-10)
-10 = a(-10)
1 = a
So the equation must be
y = 1(x²-3x-10)
or just
y = x²-3x-10
You can put this in vertex form to give a reason why it
also has vertex (1.5,-12)
y = x²-3x-10
Add 10 to both sides to get the -10 off the right side:
y+10 = x²-3x
Complet the square on the right
1. Multiply the coefficient of x by 1/2, getting -3(1/2) = -1.5\
2. Square this value (-1.5)² = 2.25
3. Add that to both sides,
y+10+2.25 = x²-3x+2.25
y+12.25 = (x-1.5)(x-1.5)
y+12.25 = (x-1.5)²
y = (x-1.5)²-12.25
Compare to
y = a(x-h)²+k
So the vertex is (1.5,-12.25).
You must have meant that the vertex is (1.5,-12.25) not (1.5,12).
Here is the graph of the parabola with x-intercepts at 5 and -2,
y-intecept at -10, and vertex (turning point at (1.5,-12.25)
So you see it crosses the x-axis at 5 and -2 and crosses the y-axis
at -10. However it has vertex (turning point) (1.5,-12.25).
So it had to be 12.25, not 12.
Edwin
