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.

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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.

Algebra
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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.
i have no clue how to do that
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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this the the graph
1 Attachment
i still dont understand
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
kropot you there?
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.
can you help me find the x and y intercepts
Put y equal to zero. Then use the formula for solving any quadratic to find the two values of x.
and that is the equation for that parabola using the 3 points i said above?
i dont know how to use the quadratic formula, very well
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: \[\large -x^{2}+10x-23\]
i still have no clue how to find them
"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 :)
so -23 would be my y-intercept correct?
To find point C on the graph you need to find the value of the quadratic when x = 5.
and how exactly do i do that?
Yes, the intercept on the y-axis is at y = -23.
so now that i have my y-intercept how do i find my x-intercept?
Solve the quadratic \[\large -x^{2}+10x-23=0\] using \[\large x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]
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 \[\large y=-x^{2}+10x-23\]
so the x-intercept is 2?
when i pulgged x into the equation i got y=-5^2+10(5)-23
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.
so am i right about the x-inttercept?
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.
but thats what i got when i plugged it into the equation
The intercepts on the x-axis are found from \[\large x=\frac{-10\pm \sqrt{8}}{-2}\]
the answer i got is definetly not right
i got 5+-sqrt2
What is the answer then?
5?
im not sure, im really bad at math
How do you know that the result for the intercepts on the x-axis being 5 +-square root 2 is not right?
That result checks out for me.
so its right?
Yes.
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
but the way that says it makes it seem like it should be points but im not sure
Try the values 3.586 and 6.414 for the intercepts on the x-axis.
and -23 for the y?
Yes.
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).
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?
You're welcome. Please post any more questions on OS. You can tag me if you like.
okay thank you!
If this is answered, close the question, please. :)
You're welcome :)

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