quadratic equation math help.?

Question

hii ok so i have another homework question..

- use the graphs to determine the equation of the qudratic functions in standard form.

its a parabola opening down, the roots are at -4 and 7. and they give me another point which is (6,32). there is no vertex point given.

i started by doing this :

a(x+4)(x-7) =0
a(x^2-7x+4x-28)=0
a(x^2-3x-28)=0

then i plugged in the points to find a.

32=0(36-18-28)
32=(a)(-10)

NOW, here's where the problem is..

a is supposed to equal -3.2. (the answer is -3.2x^2+9.6+89.6).

but how is 32*10 = -3.2??? :S :S

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also, how do you find the equation when its a parabola opening up. they only give you the vertex (3,5) and one point on the graph which is (9,-35) ??

please include the steps in your answer plz!! :) thx!

oops sry i made a typo, its supposed to be

32=a(36-18-28) not 32=0(36-18-28)

thx!! but sry, the answer says its f(x)= 5/6x^2 -5x + 25/2 ?

Answer 1

32=(a)(-10)

Divide both sides by -10:

32/(-10) = a(-10)/(-10)
-3.2 = a

There's no problem there, you got the right answer. Looks like you just multiplied by 10 instead of dividing, and lost the minus sign, when trying to simplify.

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The vertex of the parabola at (a,b) means that the parabola's equation is of the form:

c(x-a)^2 + b = y

... you substitute the given a and b, and then plug in the other point to solve for c.

c(x-3)^2 + 5 = y
y = c(x^2 - 6x + 9) + 5

The other point is (9,-35):

-35 = c(9^2 - 6*9 + 9) + 5
-40 = c(81 - 54 + 9)
-40 = c(36)
-40/36 = c
-10/9 = c

Now that you know c, plug it in to simplify:

y = c(x^2 - 6x + 9) + 5
y = -10/9 (x^2) - (-10/9)(6x) + (-10/9)(9) + 5
y = -10/9 x^2 + 60/9 x - 10 + 5
y = -10/9 x^2 + 20/3 x - 5

Test it out, try (3,5):

y = -10/9 x^2 + 20/3 x - 5
5 =? -10/9 * (3^2) + 20/3 * (3) - 5
5 =? -10 + 20 - 5
5 = 5

Also try the other point (9,-35):

y = -10/9 x^2 + 20/3 x - 5
-35 =? -10/9 * (9^2) + 20/3 * 9 - 5
-35 =? -90 + 60 - 5
-35 = -35

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You gave a different answer, which I think corresponds to the other point being (9,35), not "negative 35." That solution picks up where we calculate c:

The other point is (9,35):

35 = c(9^2 - 6*9 + 9) + 5
30= c(81 - 54 + 9)
30 = c(36)
30/36 = c
5/6 = c

Now that you know c, plug it in to simplify:

y = c(x^2 - 6x + 9) + 5
y = 5/6 (x^2) - (5/6)(6x) + (5/6)(9) + 5
y = 5/6 x^2 - 5 x + 15/2 + 5
y = 5/6 x^2 - 5 x + 25/2

Test it out, try (3,5):

y = 5/6 x^2 - 5 x + 25/2
5 =? 5/6 * (3^2) - 5*3 + 25/2
5 =? 15/2 - 15 + 25/2
5 =? 20 - 15
5 = 5

Also try the other point (9,35):

y = 5/6 x^2 - 5 x + 25/2
35 =? (5/6)*(9^2) - 5*9 + 25/2
35 =? (5/6)*81 - 45 + 25/2
35 =? 135/2 + 25/2 - 45
35 =? 80 - 45
35 = 35

Answer 2

You have 32 = a*-10. So a = -3.2. You know whether it opens up or down by the sign of the coefficient of the quadratic term (in this case -3.2). If it is positive, the parabola will open up and if it is negative, the parabola will be upside down and open downward.

Answer 3

the 1st 2 are in basic terms trial and blunder, in this typical occasion: x^2 + bx + c (x + p) (x + q) p and q could situations at the same time to furnish c and p and q could upload to furnish b. x^2+3x-10=0 (x - 2) (x + 5) = 0 x^2-5x-14=0 (x - 7) (x + 2) = 0 it truly is the version of two squares: x^2 - y^2 = (x - y) (x + y) x^2-9 = (x - 3) (x + 3)