1. Equation of function

Can u help me by explaining and show working out 2 this question? I can understand when some1 explains and shows me.....I've attempted it but cant

a) Determine the equation of the function whose graph is shown, given that it has the form y=ax^2 + bx + c, for -6<x<2.

points on graph are; (-6,13) (2,5)
crosses y axis at (0,1)
b) Determine the turning point of the curve.

THIS ISNT FOR AN ASSIGNMENT OR EXAM THING. Just normal work.

2. where is the graph?

3. Originally Posted by wandering_lost
a) Determine the equation of the function whose graph is shown, given that it has the form y=ax^2 + bx + c, for -6<x<2.

points on graph are; (-6,13) (2,5)
crosses y axis at (0,1)
Substitute the three points to get a system of linear equations,

$\left\{\begin{array}{rcrcrcr}
13&=&36a&-&6b&+&c\\
5&=&4a&+&2b&+&c\\
1&=&&&&&c
\end{array}\right.$

Solve for $a,\;b,$ and $c.$ You should get $(a,\,b,\,c)=\left(\frac12,\,1,\,1\right).$

b) Determine the turning point of the curve.
Set the derivative equal to zero, solve, and test the intervals to see where the derivative changes sign.

4. Originally Posted by wandering_lost
Can u help me by explaining and show working out 2 this question? I can understand when some1 explains and shows me.....I've attempted it but cant

a) Determine the equation of the function whose graph is shown, given that it has the form y=ax^2 + bx + c, for -6<x<2.

points on graph are; (-6,13) (2,5)
crosses y axis at (0,1)
b) Determine the turning point of the curve.

THIS ISNT FOR AN ASSIGNMENT OR EXAM THING. Just normal work.
Part A.
Use the y-intercept (0,1) to find c.

y = ax^2 + bx + c
1 = a*0 + b * 0 + c
c = 1

Now use one of the other points to find a relation between a and b. I'll use (2,5)

y = ax^2 + bx + 1
5 = a * 2^2 + b * 2 + 1
4 = 4a + 2b
2b = 4 - 4a
b = (4 - 4a)/2
b = 2 - 2a

Now replace b in the original formula with the relation you just found.

y = ax^2 + bx + 1
y = ax^2 + (2 - 2a) * x + 1

Now use that equation and the last point (-6,13) to solve for a.

y = ax^2 + (2 - 2a) * x + 1
13 = a * -6^2 + (2 - 2a) * -6 + 1
13 = a * 36 - 12 + 12a + 1
24 = 36a + 12a
48a = 24
a = 1/2 = 0.5

Now plug a back into the original equation and use any point to find b. I'll use (2,5)

y = ax^2 + bx + 1
y = 0.5 * x^2 + bx + 1
5 = 0.5 * 2^2 + b * 2 + 1
5 = 0.5 * 4 + 2b + 1
2 = 2b
b = 1

Now plug a, b, and c into the original equation

y = ax^2 + bx + c
y = 0.5x^2 + x + 1

Part B
Divide by 0.5
y = 0.5x^2 + x + 1
2y = x^2 + 2x +2

Complete the square
2y = x^2 + 2x + 2
2y = x^2 + 2x + 2 + 1 - 1
2y = x^2 + 2x + 1 + 1
2y = (x + 1)^2 + 1

Divide by 2
2y = (x + 1)^2 + 1
y = 0.5(x + 1)^2 + 0.5

From this equation the x-value at the turning point is -1 and the y-value at the turning point is 0.5. Therefore, (-1,0.5)