SOLUTION: How do you find real numbers a,b and c so that the graph of the function {{{ y=ax^2+bx+c }}} contains the points (-1,5), (3,8) and (0,2)? My process went as (-1,5) is a poin

Algebra ->  Linear-equations -> SOLUTION: How do you find real numbers a,b and c so that the graph of the function {{{ y=ax^2+bx+c }}} contains the points (-1,5), (3,8) and (0,2)? My process went as (-1,5) is a poin      Log On


   

Question 1080078: How do you find real numbers a,b and c so that the graph of the function +y=ax%5E2%2Bbx%2Bc+ contains the points (-1,5), (3,8) and (0,2)?
My process went as
(-1,5) is a point on the graph then values are, x=-1 and y=5 that must satisfy y=ax^2+bx+c
Equation one being +a%28-1%29%5E2%2Bb%28-1%29%2Bc=5+
(3,8) is a point on the graph then values are, x=3 and y=8 that must satisfy y=ax^2+bx+c
Equation two being +a%283%29%5E2%2Bb%283%29%2Bc=8+
(0,2) is a point on the graph then values are, x=0 and y=2 that must satisfy y=ax^2+bx+c
Equation three being +a%280%29%5E2%2Bb%280%29%2Bc=2+
that gives the system of 3 equation which are
ONE +a-b%2Bc=5+
TWO +9a%2B3b%2Bc=8
THREE +c=2+
then you start with ONE and THREE
+a%5E2-b%2Bc=5+
+c=2+ < multiply each side by -1
and add to get a-b=3
Im not sure if thats the way to go where you then use equation 3 to eliminate c from equation 2 or if there's another process to get solutions of a, b and c? I'm thankful for any advice on this problem!
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
First use the point of one of the given roots, (0,2).
y=a%2A0%5E2%2Bb%2A0%2Bc
2=0%2B0%2Bc
highlight%28c=2%29

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Make two equations with each of the other given points.
ax%5E2%2Bbx=y-c
ax%5E2%2Bbx=y-2
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a%2A%28-1%29%5E2%2Bb%2A%28-1%29=5-2
a-b=3
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a%2A3%5E2%2Bb%2A3=8-2
9a%2B3b=6, simplifiable,
3a%2Bb=2
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Now solve this system: system%28a-b=3%2C2a%2Bb=2%29
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.
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Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

How do you find real numbers a,b and c so that the graph of the function +y=ax%5E2%2Bbx%2Bc+ contains the points (-1,5), (3,8) and (0,2)?
My process went as
(-1,5) is a point on the graph then values are, x=-1 and y=5 that must satisfy y=ax^2+bx+c
Equation one being +a%28-1%29%5E2%2Bb%28-1%29%2Bc=5+
(3,8) is a point on the graph then values are, x=3 and y=8 that must satisfy y=ax^2+bx+c
Equation two being +a%283%29%5E2%2Bb%283%29%2Bc=8+
(0,2) is a point on the graph then values are, x=0 and y=2 that must satisfy y=ax^2+bx+c
Equation three being +a%280%29%5E2%2Bb%280%29%2Bc=2+
that gives the system of 3 equation which are
ONE +a-b%2Bc=5+
TWO +9a%2B3b%2Bc=8
THREE +c=2+
then you start with ONE and THREE
+a%5E2-b%2Bc=5+
+c=2+ < multiply each side by -1
and add to get a-b=3
Im not sure if thats the way to go where you then use equation 3 to eliminate c from equation 2 or if there's another process to get solutions of a, b and c? I'm thankful for any advice on this problem!
Great job in getting up to that point.

a - b + c = 5 ----- eq (i)

9a + 3b + c = 8 ----- eq (ii)

c = 2 ------ eq (iii)

Now, just substitute 2 for c in eq (i) to get: a – b = 3 ------- eq (iv)

Then, substitute 2 for c in eq (ii) to get: 9a + 3b = 6, which reduces to 3a + b = 2 ------- eq (v)



Now, ADD eqs (v) & (iv) to ELIMINATE b

From there, you should be able to find the values of a and b, you already have the value of c, so you have everything to form the equation.