# TP of quadratic when given y & x ints

• February 24th 2006, 06:40 PM
freswood
TP of quadratic when given y & x ints
A parabola has a turning point at (z, -8); it intersects the y-axis at y=10 and one of the x-intercepts is x=5.

Find the value of z
Find the equation of the parabola

It's seemingly easy but I can't seem to do it.
• February 24th 2006, 09:29 PM
earboth
Quote:

Originally Posted by freswood
A parabola has a turning point at (z, -8); it intersects the y-axis at y=10 and one of the x-intercepts is x=5.

Find the value of z
Find the equation of the parabola

It's seemingly easy but I can't seem to do it.

Hello,

if the vertex of the parabola has the coordinates (z, v) then the equation of the parabola, which you should use, is:
$y=a \cdot(x-z)^2+v$

You've got 3 points, which belong to the parabola:
P(0, 10)
V(z, -8)
N(5, 0)

Put in the coordinates of these points into the equation and you'll get:
$10=az^2$
$0=a\cdot(5-z)^2-8$

Solve the 1rst equation for a: $a=\frac{18}{z^2}$
and put it into the 2nd equation. You'll get:
$0=\frac{18}{z^2} \cdot (5-z)^2-8$

Solve for z. The results are: z=3 or z=15

I've attached a graph to demonstrate these results.

Greetings

EB
• February 24th 2006, 09:52 PM
ticbol
Quote:

Originally Posted by freswood
A parabola has a turning point at (z, -8); it intersects the y-axis at y=10 and one of the x-intercepts is x=5.

Find the value of z
Find the equation of the parabola

It's seemingly easy but I can't seem to do it.

Here is one way.

The turning point---the vertex--- is at (z,-8) and the parabola intersects the x-axis in two points, and it intersects the y-axis at one point only. That means the parabola is a vertical parabola---it opens upward or downward.

The standard form of the equation of a vertical parabola whose vertex is at (h,k) is:
(y-k) = a(x-h)^2 ----***

Given: two points (0,10) and (5,0), and vertex (z,-8)

At point (0,10),
(10 -(-8)) = a(0 -z)^2
18 = a(z^2) ----------------(1)

At point (5,0),
(0 -(-8)) = a(5 -z)^2
8 = a(5 -z)^2 --------------(2)

From (1), a = 18/(z^2), substitute that into (2),
8 = [18/(z^2)][25 -10z +z^2]
Clear the fractions, multiply both sides by z^2,
8z^2 = 18[25 -10z +z^2]
Divide both sides by 2,
4z^2 = 9[25 -10z +z^2]
4z^2 = 225 -90z +9z^2
0 = 225 -90z +9z^2 -4z^2
0 = 225 -90z +5z^2
Divide both sides by 5,
0 = 45 -18z +z^2
Or,
z^2 -18z +45 = 0
(z -15)(z -3) = 0
z = 15 or 3 -----------------answer.

Since there are two possible z's, then there are two possible parabolas.

One is using the vertex (15,-8).
First we find the "a" here.
Substitute z=15 into (1),
18 = a(15^2)
a = 18/225
So,
(y -(-8)) = (18/225)(x -15)^2
y +8 = (18/225)(x -15)^2 -----------answer.
If you want to expand that, and get the general form---y = ax^2 +bx +c---of the parabola,
y +8 = (18/225)(x^2 -30x +225)
y +8 = (18/225)x^2 -(540/225)x +18
y = (18/225)x^2 -(12/5)x +18 -8
y = 0.08x^2 -2.4x +10 ------------------answer.

The other is using the vertex (3,-8).
a = 18/(3^2) = 2
So,
(y -(-8)) = 2(x -3)^2
y +8 = 2(x -3)^2 -------------answer.
Or,
y +8 = 2(x^2 -6x +9)
y +8 = 2x^2 -12x +18
y = 2x^2 -12x +18 -8
y = 2x^2 -12x +10 -----------answer.
• February 24th 2006, 11:13 PM
freswood
Thanks heaps for your help! It makes sense now. Stupid me - I always rely on elimination, and never think to use substitution :D