help with quadratics!! (1 Viewer)

[jjuunnee]

Question:

The general form of a quadratic with zeroes x=2 and x=8 is y=a(x-2)(x-8). Find the equation of such a quadratic for which:

a) the coefficient of x² is 3
b) the constant term is -3
c) the y intercept is -16
d) the coefficient of x is 25
e) the vertex is (5,-12)
f) the curve passes through (1,-20)

Not sure how to answer these questions so explanations would be great. Please dont just answer them. Thanks!

 

[WildestDreams]

a) Well for this part the answer is clearly a=3. Think about, if you were to expand (x-2)(x-8) you'd be left with a monic result, x^2 and now all you need to do is times it by 3 to get your answer.

b) x^2 -10x +16. Divide by 16 and multiply by 3. Therefore a=3/16

c) a= -1. Multiply everything by negative 1 to get the opposite sign.

d) a = 5/3 *** Not sure about this one someone else check!

e) Sub values in formula to get value of a

y = a(x-2)(x-8)
-12 = a(5-2)(5-8)
-12 = a x -9
Therefore a = 4/3

f) Use same method

 

[integral95]

if you expand the quadratic, you get

ax^2+10ax+16a

so for part a) the answer is the equation with a = 3, as the coefficient of x^2 is the term in front of it.
use this expression for the remaining parts.

 

[InteGrand]

Question:

The general form of a quadratic with zeroes x=2 and x=8 is y=a(x-2)(x-8). Find the equation of such a quadratic for which:

a) the coefficient of x² is 3
b) the constant term is -3
c) the y intercept is -16
d) the coefficient of x is 25
e) the vertex is (5,-12)
f) the curve passes through (1,-20)

Not sure how to answer these questions so explanations would be great. Please dont just answer them. Thanks!

For each of these, we just need to find the appropriate value of a. Here are some hints:

a) What is the coefficient of x^2 of the given quadratic in terms of a?
b) What is the constant term in terms of a? (Product of roots times a)
c) The y-intercept is found by subbing x = 0 into the given equation. It will be in terms of a. Set this equal to -16 and solve for a.
d) What is the x coefficient of the given quadratic in terms of a (it is -a*(sum of roots); expand the quadratic if it's easier for you that way)?
e) The vertex of the quadratic has x-value 5 due to what the roots are. How would we then find the y-value of the quadratic in terms of a?
f) Sub. the given point into the given equation and solve for a.

 

[boredofstudiesuser1]

Question:

The general form of a quadratic with zeroes x=2 and x=8 is y=a(x-2)(x-8). Find the equation of such a quadratic for which:

a) the coefficient of x² is 3
b) the constant term is -3
c) the y intercept is -16
d) the coefficient of x is 25
e) the vertex is (5,-12)
f) the curve passes through (1,-20)

Not sure how to answer these questions so explanations would be great. Please dont just answer them. Thanks!

To answer this question (as a whole), I would begin by expanding the equation they gave you to know what you have to sub in. This would give you:

y=a(x²-10x+16)
=ax²-10ax+16a

This allows you to answer the questions having a better idea of what you're finding, which is a. Then to find the equation you just sub a in.

Keep in mind that in d, when I use the formula -b/a, I don't mean the a we're finding (although it ends up being that) but I mean a as in what is the co-efficient of the term x² just as b in the x term and c is the constant.

a) For the co-efficient of x² to be 3, you would have to make the value of whatever is in front of the x² term, 3. In this case, since a is there, replace ^a for 3, this would give you.

y=(3)x²-10(3)x+16(3)
=3x²-30x+48

NOTE: The x term and constant did also change but it doesn't matter as the only requirement for this was for the co-efficient of x² to be 3, which we have achieved.

b) Now as we can see that finding a gives us what we need, we can solve this one easily.

We need to turn the current constant term (16a) into -3. So we can look at it this way:

16a=-3

Then solve that:

a=-3/16

Then sub -3/16 as a into the original equation.

y=(-3/16)x²-10(-3/16)x+16(-3/16)
=-3/16x²+30/16x+(-48/16)
=-3/16x²+30/16x-3

c) Same method as b

16a=-16
a=-1

Then sub into original equation:

y=-x²+10x-16

Basically, all the signs just changed.

d) Same method as b and c, except using the x term.

-10ax=25x
-10a=25
a=-25/10
=-2.5

Sub a into the equation:

y=(-2.5)x²-10(-2.5)x+16(-2.5)
=-2.5x²+25x-40

e) To find the vertex of a parabola, you use a formula and then substitution. Usually you would find x by doing -b/a. Since in this case we have x, we need to find -b/a. Then to find y you sub the x value into the equation. Since we know the x and y value but not a and b, we can work in reverse. So we can just sub in x to "find y", knowing that y needs to equal -12, and we do an equation to find a:

a(5)²-10a(5)+16a = -12

25a-50a+16a=-12
-9a=-12
a=-12/-9
=12/9

Then that means the value of a would be 12/9 which = 4/3.

Sub this in to the original equation:

y=4/3x²-10(4/3)x+16(4/3)
=4/3x²-40/3x+64/3

I am not fully sure about this one, this is just the method I would use, someone can double check this.

f) Uses same method as e.

a-10a+16a=-20
7a=-20
a=-20/7

y=-20/7x²-10(20/7)x+16(10/7)
=-20/7x²-200/7x+160/7

Hope this helps and was what you were looking for.