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anonymous
 one year ago
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
 one year ago
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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kropot72
 one year ago
Best ResponseYou've already chosen the best response.2First 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
 one year ago
Best ResponseYou've already chosen the best response.0i have no clue how to do that

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2To 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i still dont understand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2If 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
 one year ago
Best ResponseYou've already chosen the best response.0can you help me find the x and y intercepts

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2Put y equal to zero. Then use the formula for solving any quadratic to find the two values of x.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and that is the equation for that parabola using the 3 points i said above?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont know how to use the quadratic formula, very well

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2By 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: \[\large x^{2}+10x23\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i still have no clue how to find them

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2"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
 one year ago
Best ResponseYou've already chosen the best response.0so 23 would be my yintercept correct?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2To find point C on the graph you need to find the value of the quadratic when x = 5.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and how exactly do i do that?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2Yes, the intercept on the yaxis is at y = 23.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now that i have my yintercept how do i find my xintercept?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2Solve the quadratic \[\large x^{2}+10x23=0\] using \[\large x=\frac{b \pm \sqrt{b^{2}4ac}}{2a}\]

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2To 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 \[\large y=x^{2}+10x23\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the xintercept is 2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when i pulgged x into the equation i got y=5^2+10(5)23

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.0so am i right about the xinttercept?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2The intercepts on the xaxis are found by using the formula for solving any quadratic and putting y = 0. The xintercept is not 2.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but thats what i got when i plugged it into the equation

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2The intercepts on the xaxis are found from \[\large x=\frac{10\pm \sqrt{8}}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the answer i got is definetly not right

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2What is the answer then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im not sure, im really bad at math

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2How do you know that the result for the intercepts on the xaxis being 5 +square root 2 is not right?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2That result checks out for me.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Once 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, rightclick 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+10x23 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 yintercepts 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 xintercepts, select the parabola, then the xaxis. To find the yintercepts, select the parabola, then the yaxis. Xintercepts: (,) Yintercepts: (23) this is what im stuck on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the way that says it makes it seem like it should be points but im not sure

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2Try the values 3.586 and 6.414 for the intercepts on the xaxis.

kropot72
 one year ago
Best ResponseYou've already chosen the best response.2The intercepts on the xaxis are at (3.586, 0) and (6.414, 0) and the intercept on the yaxis is at (0, 23).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay 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
 one year ago
Best ResponseYou've already chosen the best response.2You're welcome. Please post any more questions on OS. You can tag me if you like.

hlambach
 one year ago
Best ResponseYou've already chosen the best response.0If this is answered, close the question, please. :)
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