Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph.

f(x)=x^2-3
point: (-5,22)

Question

Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph.

f(x)=x^2-3
point: (-5,22)

michele_laino · Answer

the requested slope, is the slope of the tangent line at point $(-5,22)$. More precisely, it is the first derivative of the function, at $x=-5$

whpalmer4 · Answer

Find the first derivative of $f(x)$, and evaluate it at $x=-5$ to find the slope at $(-5,22)$. 

Then find the equation of a line with that slope passing through that point, using slope-intercept form.

michele_laino · Answer

so, what is the first derivative of the function:
$F(x)=x^2-3$

anonymous · Answer

Is it 2x?

zepdrix · Answer

yes

anonymous · Answer

so the slope is -10?

whpalmer4 · Answer

Yes, the slope will be $-10$ at $(-5,22)$.  Did you find the equation for the line?