Question

find the slope of the tangent line to the graph of the function at the given point.
$f(x) = 4 - 3x, \\ (-1, 7)$

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1. -/0.5 points

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find a formula for the slope $m$ of the tangent line to $f(x) = x^2 + 4x$ at any point $x = a$.

Explanation:

Step1: Differentiate $f(x)=4-3x$

The derivative of a linear function $f(x)=mx+b$ is $f'(x)=m$. For $f(x)=4-3x$, $f'(x) = -3$.

Step2: Slope is constant for linear functions

For any point on a linear function, the slope of the tangent line equals the function's slope. So at $(-1,7)$, the slope is $-3$.

Step3: Differentiate $f(x)=x^2+6x$

Use power rule: $\frac{d}{dx}(x^n)=nx^{n-1}$.
$f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(6x) = 2x + 6$

Step4: Substitute $x=a$ for slope formula

Replace $x$ with $a$ in $f'(x)$: $m = 2a + 6$.

Answer:

1. $-3$
2. $2a + 6$