# Given the quadratic function y=ax^2+bx+c, the maximum value is a^2+ 4, at x =1, and the graph passes through point (3, 1). How do you find the values of the constants a, b, c?

##### 1 Answer
Jan 12, 2017

$\left(a , b , c\right) = \left(- 3 , 6 , 10\right) , \mathmr{and} , \left(- 1 , 2 , 4\right) .$

#### Explanation:

The graph of the function (fun.) $y = a {x}^{2} + b x + c$ passes through

the point (pt.) $\left(3 , 1\right)$, so, the co-ordinates must satisfy the

equation.

$\therefore 1 = 9 a + 3 b + c \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$

$y \text{ is maximum at } x = 1 \therefore {\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{x = 1} = 0 \therefore 2 a + b = 0. \ldots . . \left(2\right)$

Next,

${y}_{\max} = {a}^{2} + 4 \text{ occurs at } x = 1 \therefore a + b + c = {a}^{2} + 4. \ldots \ldots \left(3\right)$

(1),(2), &,(3) rArr (a,b,c)=(-3,6,10), or, (-1,2,4).

For ${y}_{\max} \text{ at x=1, we must have, } {\left[\frac{{d}^{2} y}{\mathrm{dx}} ^ 2\right]}_{x = 1} < 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 a x + b \Rightarrow \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 2 a = - 6 \mathmr{and} - 2 , \text{ readily } < 0.$

It is easy to verify that the pair of triads so derived satisfies the given

conditions.

$\text{Therefore, } \left(a , b , c\right) = \left(- 3 , 6 , 10\right) , \mathmr{and} , \left(- 1 , 2 , 4\right) .$

Enjoy Maths.!