The graph of y=ax^2+bx has an extremum at (1,-2). Find the values of a and b?

May 23, 2018

$a = 2$ and $b = - 4$

Explanation:

Given: $y = a {x}^{2} + b x , y \left(1\right) = - 2$

From the given can substitute 1 for x and 2 for y and write the following equation:

$- 2 = a + b \text{ [1]}$

We can write the second equation using that the first derivative is 0 when $x = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 a x + b$

$0 = 2 a + b \text{ [2]}$

Subtract equation [1] from equation [2]:

$0 - - 2 = 2 a + b - \left(a + b\right)$

$2 = a$

$a = 2$

Find the value of b by substituting $a = 2$ into equation [1]:

$- 2 = 2 + b$

$- 4 = b$

$b = - 4$

May 23, 2018

$f \left(x\right) = 2 {x}^{2} - 4 x$

Explanation:

$f \left(x\right) = a {x}^{2} + b x$ , $x$$\in$$\mathbb{R}$

• $1$$\in$$\mathbb{R}$
• $f$ is differentiable at ${x}_{0} = 1$
• $f$ has an extremum at ${x}_{0} = 1$

According to Fermat's Theorem $f ' \left(1\right) = 0$

but $f ' \left(x\right) = 2 a x + b$

$f ' \left(1\right) = 0$ $\iff$ $2 a + b = 0$ $\iff$ $b = - 2 a$

$f \left(1\right) = - 2$ $\iff$ $a + b = - 2$ $\iff$ $a = - 2 - b$

So $b = - 2 \left(- 2 - b\right)$ $\iff$ $b = 4 + 2 b$ $\iff$

$b = - 4$

and $a = - 2 + 4 = 2$

so $f \left(x\right) = 2 {x}^{2} - 4 x$