Find the slope of the function's graph at the \( g \) \( f(x)=x^{2}+1,(-5,26) \) What is the slope of the function's graph at th \( m=-10 \) (Simplify your answer.) Find an equation for the line tangent to the gi \( y=\square \)

To find the slope of the function \( f(x) = x^2 + 1 \) at the point \( (-5, 26) \), we first need to calculate the derivative \( f'(x) \). The derivative is given by \( f'(x) = 2x \). Now, plug in \( x = -5 \) into the derivative to get the slope: \[ f'(-5) = 2(-5) = -10. \] So, the slope of the tangent line at that point is \( -10 \). Now, to find the equation of the tangent line, we can use the point-slope form of the equation of a line, which is \( y - y_1 = m(x - x_1) \). Here, \( m = -10 \), \( x_1 = -5 \), and \( y_1 = 26 \): \[ y - 26 = -10(x + 5). \] Simplifying this yields: \[ y - 26 = -10x - 50 \implies y = -10x - 24. \] Thus, the equation of the tangent line is \( y = -10x - 24 \). Fun Fact: Slope in calculus isn’t just a number—it’s a representation of how fast something is changing! When a car travels on a hilly road, the slope indicates whether it’s climbing up a hill or racing down one! Practical Tip: When finding the equation of a tangent line, always remember to use the point-slope form right after calculating the slope. It’s a straightforward method that saves you from making any calculation errors!