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.

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- Feb 24th 2006, 06:40 PMfreswoodTP 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. - Feb 24th 2006, 09:29 PMearbothQuote:

Originally Posted by**freswood**

if the vertex of the parabola has the coordinates (z, v) then the equation of the parabola, which you should use, is:

$\displaystyle y=a \cdot(x-z)^2+v$

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:

$\displaystyle 10=az^2$

$\displaystyle 0=a\cdot(5-z)^2-8$

Solve the 1rst equation for a: $\displaystyle a=\frac{18}{z^2}$

and put it into the 2nd equation. You'll get:

$\displaystyle 0=\frac{18}{z^2} \cdot (5-z)^2-8$

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

I've attached a graph to demonstrate these results.

Greetings

EB - Feb 24th 2006, 09:52 PMticbolQuote:

Originally Posted by**freswood**

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. - Feb 24th 2006, 11:13 PMfreswood
Thanks heaps for your help! It makes sense now. Stupid me - I always rely on elimination, and never think to use substitution :D