anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
kropot72
  • 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
  • anonymous
i have no clue how to do that
kropot72
  • 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
  • anonymous
this the the graph
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anonymous
  • anonymous
i still dont understand
anonymous
  • 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
  • anonymous
kropot you there?
kropot72
  • 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
  • anonymous
can you help me find the x and y intercepts
kropot72
  • kropot72
Put y equal to zero. Then use the formula for solving any quadratic to find the two values of x.
anonymous
  • anonymous
and that is the equation for that parabola using the 3 points i said above?
anonymous
  • anonymous
i dont know how to use the quadratic formula, very well
kropot72
  • 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
  • anonymous
i still have no clue how to find them
kropot72
  • 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
  • anonymous
so -23 would be my y-intercept correct?
kropot72
  • kropot72
To find point C on the graph you need to find the value of the quadratic when x = 5.
anonymous
  • anonymous
and how exactly do i do that?
kropot72
  • kropot72
Yes, the intercept on the y-axis is at y = -23.
anonymous
  • anonymous
so now that i have my y-intercept how do i find my x-intercept?
kropot72
  • kropot72
Solve the quadratic \[\large -x^{2}+10x-23=0\] using \[\large x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]
kropot72
  • 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
  • anonymous
so the x-intercept is 2?
anonymous
  • anonymous
when i pulgged x into the equation i got y=-5^2+10(5)-23
kropot72
  • 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
  • anonymous
so am i right about the x-inttercept?
kropot72
  • 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
  • anonymous
but thats what i got when i plugged it into the equation
kropot72
  • kropot72
The intercepts on the x-axis are found from \[\large x=\frac{-10\pm \sqrt{8}}{-2}\]
anonymous
  • anonymous
the answer i got is definetly not right
anonymous
  • anonymous
i got 5+-sqrt2
kropot72
  • kropot72
What is the answer then?
anonymous
  • anonymous
5?
anonymous
  • anonymous
im not sure, im really bad at math
kropot72
  • kropot72
How do you know that the result for the intercepts on the x-axis being 5 +-square root 2 is not right?
kropot72
  • kropot72
That result checks out for me.
anonymous
  • anonymous
so its right?
kropot72
  • kropot72
Yes.
anonymous
  • 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
  • anonymous
but the way that says it makes it seem like it should be points but im not sure
kropot72
  • kropot72
Try the values 3.586 and 6.414 for the intercepts on the x-axis.
anonymous
  • anonymous
and -23 for the y?
kropot72
  • kropot72
Yes.
kropot72
  • 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
  • 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
  • kropot72
You're welcome. Please post any more questions on OS. You can tag me if you like.
anonymous
  • anonymous
okay thank you!
hlambach
  • hlambach
If this is answered, close the question, please. :)
kropot72
  • kropot72
You're welcome :)

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