Finding the other X value of a quadratic

[spwittbold]

Hello,

I'm looking for some assistance with this problem - I'm not sure quite what to do.

The problem states:the x=-The graph of a parabola of the form y = x^2 + bx + c crosses the y axist at -36 and the x axis at 3. At what other point does the line cross the x axis?

So, couldn't figure out how to solve this.

The work I tried to do was use the points (0, - 36) and ((3, 0) in that equation as follows:

0 = a(3)^2 + b(3) + c
-36 = a(0) + b(0) + c

Therefore C must = -36 because of equation 2.

C = 36

0 = a(3)^2 + b(3) + -36
-36 = a(0) + b(0) + -36

We are left with:
0 = 3a+3b-36

How would you recommend I proceed? How would someone go about solving this problem?

 

[spwittbold]

Mtel #38

The graph of a parabola of the form y = x^2 +bx + c crosses the y axis at -36 and the x axis at 3. At what other point does the graph cross the x-axis?

Not sure what to do here.

Does this suggest A = 1, C = -36 and B = unknown?

This is what I did.

-36 = 0 + 0 + C, so C = -36

0 = 3^2 + b(3) + -36

0 = 9 + 3b + -36

3b = - 27

b = 9

Use original equation with point (3, 0)

0 = 3^2 + 9*3 + - 36

0 = 9 + 27 + - 36

0 = 0 is a true statement, so I've found coefficients of A B and C and A = 1, B = 9 and C = -36. So therefore the equation is y = 1x^2 + 9x -36


....and I solved my own problem because that factors to (x-3)(x+12) and x = 3 and x = -12. B ALSDH

 

[Ishuda]

Hello,

I'm looking for some assistance with this problem - I'm not sure quite what to do.

The problem states:the x=-The graph of a parabola of the form y = x^2 + bx + c crosses the y axist at -36 and the x axis at 3. At what other point does the line cross the x axis?

So, couldn't figure out how to solve this.

The work I tried to do was use the points (0, - 36) and ((3, 0) in that equation as follows:

0 = a(3)^2 + b(3) + c
-36 = a(0) + b(0) + c

Therefore C must = -36 because of equation 2.

C = 36

0 = a(3)^2 + b(3) + -36
-36 = a(0) + b(0) + -36

We are left with:
0 = 3a+3b-36

How would you recommend I proceed? How would someone go about solving this problem?

What is the value of a in the function as given
y = x^2 + bx + c = a x^2 + bx + c

 

[Deleted member 4993]

Hello,

I'm looking for some assistance with this problem - I'm not sure quite what to do.

The problem states:the x=-The graph of a parabola of the form y = x^2 + bx + c crosses the y axist at -36 and the x axis at 3. At what other point does the line cross the x axis?

So, couldn't figure out how to solve this.

The work I tried to do was use the points (0, - 36) and ((3, 0) in that equation as follows:

0 = a(3)^2 + b(3) + c
-36 = a(0) + b(0) + c

Therefore C must = -36 because of equation 2.

C = 36 ..................... Correct

Now calculate the x-intercepts (roots) of the function:

y = x^2 + bx + 36

0 = 9 + 3b + 36 → b = -15

x_1,2 = [-b ± √ (b^2 - 144)]/(2) ............................ these are the roots of the function. One of these is "= 3"

Calculate the other root

0 = a(3)^2 + b(3) + -36
-36 = a(0) + b(0) + -36

We are left with:
0 = 3a+3b-36

How would you recommend I proceed? How would someone go about solving this problem?

Let us know if you are still stuck.

 

[spwittbold]

What is the value of a in the function as given
y = x^2 + bx + c = a x^2 + bx + c

They don't provide a value for A which is why I think I was confused with the problem.

It simply says in the form of....."y = x^2 + bx + c...." so I assume you just presume a is equal to 1?

 

[Harry_the_cat]

The graph of a parabola of the form y = x^2 +bx + c crosses the y axis at -36 and the x axis at 3. At what other point does the graph cross the x-axis?

Not sure what to do here.

Does this suggest A = 1, C = -36 and B = unknown?

This is what I did.

-36 = 0 + 0 + C, so C = -36

0 = 3^2 + b(3) + -36

0 = 9 + 3b + -36

3b = - 27 .... should be 3b = 27

b = 9

Use original equation with point (3, 0)

0 = 3^2 + 9*3 + - 36

0 = 9 + 27 + - 36

0 = 0 is a true statement, so I've found coefficients of A B and C and A = 1, B = 9 and C = -36. So therefore the equation is y = 1x^2 + 9x -36


....and I solved my own problem because that factors to (x-3)(x+12) and x = 3 and x = -12. B ALSDH

see red comment

 

[Otis]

The graph of a parabola of the form y = x^2 + bx + c crosses the y axis at -36 ...

Does this suggest [a] = 1, [c] = -36 and = unknown?

Yes.

 

[Ishuda]

They don't provide a value for A which is why I think I was confused with the problem.

It simply says in the form of....."y = x^2 + bx + c...." so I assume you just presume a is equal to 1?

Yes since x^2 = 1 x^2 = a x^2 and a=1.