If $y=f(x)$ has a turning point when $x=\frac{1}{4}$, can we find $\lambda$?

If $y=f(x)$ has a turning point when $x=\frac{1}{4}$, can we find $\lambda$?

Question

It is given that $f(x)=(x-2)^2-\lambda(x+1)(x+2).$

1. Find the values of $\lambda$ for which the equation $f(x)=0$ has two equal roots.

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1. Show that, when $\lambda=2$, $f(x)$ has a maximum value of $25$.

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1. Given that the curve $y=f(x)$ has a turning point when $x=\dfrac{1}{4}$, find the value of $\lambda$ and sketch the curve for this value of $\lambda$.

Suggestion

Could we express $f(x)$ in the form $ax^2+bx+c$?

What is the condition for an equation to have two equal roots?

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Here the curve $y=f(x)$ is in red, while the line $x = \frac{1}{4}$ is in blue.

Solution

By expanding

$f(x)$

we obtain

$$\begin{align*} f(x) &= x^2-4x+4-\lambda(x^2+3x+2) \\ &= (1-\lambda)x^2-(4+3\lambda)x+4-2\lambda. \end{align*}$$

For the equation

$f(x)=0$

to have two equal roots, the

discriminant

must be zero. So we want to solve

$$\begin{align*} 0 &= [-(4+3\lambda)]^2-4(1-\lambda)(4-2\lambda) \\ &= 16+24\lambda+9\lambda^2-4(4-6\lambda+2\lambda^2) \\ &= \lambda^2+48\lambda \\ &= \lambda(\lambda+48), \end{align*}$$

and so $\lambda = 0$ or $\lambda=-48$.

When $\lambda=2$, we have $f(x)=-x^2-10x$.

We find the maximum of $f(x)$ by completing the square.

We can write $f(x)=-(x+5)^2+25$, and since $-(x+5)^2$ is strictly negative, we see that the maximum value of $f(x)$ is $25$, which occurs when $x=-5$.

Alternatively we could find the derivative $f'(x)$ and set this equal to zero to find the turning point.

Or we could find the roots of $f(x)=0$, and use the symmetry of the parabola.

We have $f'(x)=2(1-\lambda)x-(4+3\lambda).$

We know that $f'\left(\dfrac{1}{4}\right)=0$, since there is a turning point at $x=\dfrac{1}{4}$.

So $2(1-\lambda)\left(\frac{1}{4}\right)-4-3\lambda=0,$ and so $1-\lambda-8-6\lambda=0,$ on multiplying the equation by $2$. This simplifies to $\lambda = -1$.

For this value of $\lambda$, we have $f(x)=2x^2-x+6$.

To sketch this graph we observe that the $y$-intercept is $6$, and that $x=\dfrac{1}{4}$ gives a minimum, since the coefficient of the $x^2$ term is positive.

We calculate that $f\left(\frac{1}{4}\right)=\frac{2}{16}-\frac{1}{4}+6=\frac{2-4+96}{16}=\frac{94}{16}=\frac{47}{8}.$