# How do you find the slope of the graph f(x)=-1/2+7/5x^3 at (0,-1/2)?

Dec 28, 2016

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#### Explanation:

The slope of a function $f \left(x\right)$ for a non-specific point $x$ is given by the derivative of the function $f \left(x\right)$.

For the case $f \left(x\right) = - \frac{1}{2} + \frac{7}{5} {x}^{3}$

the derivative (using the exponent rule) is
$\textcolor{w h i t e}{\text{XXX}} f ' \left(x\right) = 3 \times \frac{7}{5} {x}^{3 - 1} = \frac{21}{5} {x}^{2}$

At the point $\left(0 , - \frac{1}{2}\right)$
$\textcolor{w h i t e}{\text{XXX}} x = 0$ (and $y = - \frac{1}{2}$, but that's irrelevant).
so the slope is
$\textcolor{w h i t e}{\text{XXX}} f ' \left(0\right) = \frac{21}{5} \cdot {\left(0\right)}^{2} = 0$