I need to find the equation ax^2+bx+c using the points (5,2) (6,1) (4,1). I have no clue how to do it. I will give medal to everyone that helps.

- anonymous

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- kropot72

First you need to use the ordered pairs to set up 3 equations. Then the equations must be solved to find the values of a, b and c.

- anonymous

i have no clue how to do that

- kropot72

To form the 3 equations, the values in the ordered pairs are plugged into the equation for the quadratic:
25a + 5b + c = 2 ........(1)
36a + 6b + c = 1 ........(2)
16a + 4b + c = 1 ........(3)
Now solve these equations to find the values of a, b and c.

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- anonymous

this the the graph

##### 1 Attachment

- anonymous

i still dont understand

- anonymous

i also need to find the x and y intercpets of that parabola along with the equation but i have no clue how to do it

- anonymous

kropot you there?

- kropot72

If we subtract equation (1) from equation (2) we can eliminate c as follows:
11a + b = -1 ...........(4)
And if we subtract equation (3) from equation (2) we get another equation in a and b as follows:
20a + 2b = 0 ............(5)
Multiplying equation (4) by 2 gives:
22a + 2b = -2 ..........(6)
Subtracting (5) from (6) we get:
2a = -2 ....................(7)
which gives a the value of -1.

- anonymous

can you help me find the x and y intercepts

- kropot72

Put y equal to zero. Then use the formula for solving any quadratic to find the two values of x.

- anonymous

and that is the equation for that parabola using the 3 points i said above?

- anonymous

i dont know how to use the quadratic formula, very well

- kropot72

By substitution in (4) we find the value of b is 10. And by substitution in (3) the value of c is found to be -23. The required equation is therefore:
\[\large -x^{2}+10x-23\]

- anonymous

i still have no clue how to find them

- kropot72

"and that is the equation for that parabola using the 3 points i said above?"
Yes, it is.
You had better do some work on the formula for solving any quadratic, and then solve it yourself. I shouldn't have to do all your work :)

- anonymous

so -23 would be my y-intercept correct?

- kropot72

To find point C on the graph you need to find the value of the quadratic when x = 5.

- anonymous

and how exactly do i do that?

- kropot72

Yes, the intercept on the y-axis is at y = -23.

- anonymous

so now that i have my y-intercept how do i find my x-intercept?

- kropot72

Solve the quadratic
\[\large -x^{2}+10x-23=0\]
using
\[\large x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]

- kropot72

To find point C on the graph you need to find the value of the quadratic when x = 5.
Just plug the value x = 5 into
\[\large y=-x^{2}+10x-23\]

- anonymous

so the x-intercept is 2?

- anonymous

when i pulgged x into the equation i got y=-5^2+10(5)-23

- kropot72

The ordered pair for point C on the graph is (5, 2). Just look at the graph to see that it is the coordinates for the highest point.

- anonymous

so am i right about the x-inttercept?

- kropot72

The intercepts on the x-axis are found by using the formula for solving any quadratic and putting y = 0. The x-intercept is not 2.

- anonymous

but thats what i got when i plugged it into the equation

- kropot72

The intercepts on the x-axis are found from
\[\large x=\frac{-10\pm \sqrt{8}}{-2}\]

- anonymous

the answer i got is definetly not right

- anonymous

i got 5+-sqrt2

- kropot72

What is the answer then?

- anonymous

5?

- anonymous

im not sure, im really bad at math

- kropot72

How do you know that the result for the intercepts on the x-axis being 5 +-square root 2 is not right?

- kropot72

That result checks out for me.

- anonymous

so its right?

- kropot72

Yes.

- anonymous

Once the graphed parabola matches as close as possible to the image, make note of the quadratic equation, both in general and standard form (y = ax2 + bx + c). When completed, the graphed parabola should lie on top of the parabola in your image. If necessary, right-click on the image and go to the Color tab to make the image more transparent so the graphed parabola can be seen over the image. The equation is: y=-x^2+10x-23
Image of the Preferences window with the Color tab active. A scale from 0 to 100 is shown. At 0, the image is transparent. At 100, the image is solid.
Identify the vertex, axis of symmetry, domain and range of the graphed parabola by inspecting your graph.
Vertex: (5,2)
Axis of Symmetry: x=5
Find the x- and y-intercepts of the parabola using the "Intersect Two Objects" icon image of geogebra intersect two objects button. Remember this tool may be hidden below other buttons. To find the x-intercepts, select the parabola, then the x-axis. To find the y-intercepts, select the parabola, then the y-axis.
X-intercepts: (,)
Y-intercepts: (-23)
this is what im stuck on

- anonymous

but the way that says it makes it seem like it should be points but im not sure

- kropot72

Try the values 3.586 and 6.414 for the intercepts on the x-axis.

- anonymous

and -23 for the y?

- kropot72

Yes.

- kropot72

The intercepts on the x-axis are at (3.586, 0) and (6.414, 0) and the intercept on the y-axis is at (0, -23).

- anonymous

okay thank you so much for you help, ill give you a medal. i am starting to understand it now. whats your email so i can email you if i have any other questions?

- kropot72

You're welcome. Please post any more questions on OS. You can tag me if you like.

- anonymous

okay thank you!

- hlambach

If this is answered, close the question, please. :)

- kropot72

You're welcome :)

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