Question

chapter iii.5
score: 1/3 answered: 1/3
question 2
take f to be the quadratic function
$f(x)=6x^{2}-11x + 9$.
take l to be the line that is given by $l(x)=mx + b$. identify a quadratic equation that determines m so that l is tangent to f at $(-8,481)$.

Explanation:

Step1: Find the derivative of $f(x)$

The derivative of $f(x)=6x^{2}-11x + 9$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=12x-11$.

Step2: Evaluate the slope of the tangent line at $x = - 8$

Substitute $x=-8$ into $f'(x)$. So $f'(-8)=12\times(-8)-11=-96 - 11=-107$. Also, since the line $L(x)=mx + b$ is tangent to $f(x)$ at $(-8,481)$, we know that the slope of the tangent line $m$ satisfies the following condition. The fact that the line $L(x)$ and $f(x)$ pass through the point $(-8,481)$ gives us $481=m\times(-8)+b$, or $b = 481 + 8m$. And since the line is tangent, the slope of the line $m$ should satisfy the equation based on the fact that the difference between the function and the line has a double - root at $x=-8$.

We know that if $y = f(x)-L(x)=6x^{2}-11x + 9-(mx + b)=6x^{2}-(11 + m)x+(9 - b)$ has a double - root at $x=-8$.

Substitute $x=-8$ into $y$ and set $y = 0$ and its derivative $y'$ at $x=-8$ equal to $0$. First, $y(-8)=6\times(-8)^{2}-(11 + m)\times(-8)+(9 - b)=0$. Second, $y'=12x-(11 + m)$, and $y'(-8)=12\times(-8)-(11 + m)=0$.

From $y'(-8)=-96-(11 + m)=0$, we can also work in another way.

Since the line $L(x)=mx + b$ and $f(x)$ are tangent at $(-8,481)$, we know that $f(-8)=L(-8)$ and $f'(-8)=m$.

We can also use the fact that the equation $6x^{2}-11x + 9=mx + b$ has a double - root at $x=-8$.

Substitute $x=-8$ into $6x^{2}-11x + 9=mx + b$:
\[6\times(-8)^{2}-11\times(-8)+9=m\times(-8)+b\]
\[384 + 88+9=-8m + b\]
\[481=-8m + b\]

Since the line $y = mx + b$ is tangent to $y = 6x^{2}-11x + 9$, the equation $6x^{2}-(11 + m)x+(9 - b)=0$ has a double - root. For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 6$, $b=-(11 + m)$, $c = 9 - b$), the discriminant $\Delta=b^{2}-4ac = 0$.

Another way:
Since the line $L(x)=mx + b$ is tangent to $f(x)$ at $x=-8$, we know that the function $g(x)=6x^{2}-(11 + m)x+(9 - b)$ (where $b = 481+8m$) has a double - root at $x=-8$.

We know that for a quadratic function $y = Ax^{2}+Bx + C$, if it has a double - root at $x = k$, then $y(k)=0$ and $y'(k)=0$.

The derivative of $g(x)$ is $g'(x)=12x-(11 + m)$.

Since $g'(-8)=0$, we have $12\times(-8)-(11 + m)=0$.

\[ - 96-11 - m=0\]
\[m=-107\]

We can also use the fact that the equation $6x^{2}-11x + 9-(mx + b)=0$ or $6x^{2}-(11 + m)x+(9 - b)=0$ has a double - root. The discriminant $\Delta=(11 + m)^{2}-24(9 - b)=0$. Substitute $b = 481+8m$ into it.

\[

$$\begin{align*} (11 + m)^{2}-24(9-(481 + 8m))&=0\\ 121+22m+m^{2}-24\times(-472-8m)&=0\\ 121+22m+m^{2}+11328+192m&=0\\ m^{2}+214m + 11449&=0 \end{align*}$$

\]

Answer:

$m^{2}+214m + 11449 = 0$