Do the following for the function \( g(x)=10 x^{2}+9 x \). (a) Express the slope of the secant line in terms of \( x \) and \( h \). (b) Find \( m_{s e c} \) for \( h=0.5,0.1 \), and 0.01 at \( x=1 \). What value does \( m_{s e c} \) approach as \( h \) approaches 0 ? (c) Find the equation for the secant line at \( x=1 \) with \( h=0.01 \). (d) Graph \( g \) and the secant line found in part (c) on the same viewing window. (a) Type the slope of the secant line in terms of \( x \) and \( h \). \( m_{\text {sec }}=\square \) (Simplify your answer. Do not factor.)

To find the slope of the secant line for the function \( g(x) = 10x^2 + 9x \), we first express it as follows: \[ m_{\text{sec}} = \frac{g(x+h) - g(x)}{h} \] Now, we need to compute \( g(x+h) \): \[ g(x+h) = 10(x+h)^2 + 9(x+h) = 10(x^2 + 2xh + h^2) + 9x + 9h = 10x^2 + 20xh + 10h^2 + 9x + 9h \] Now substituting back into the secant line formula, we get: \[ m_{\text{sec}} = \frac{(10x^2 + 20xh + 10h^2 + 9x + 9h) - (10x^2 + 9x)}{h} \] This simplifies to: \[ m_{\text{sec}} = \frac{20xh + 10h^2 + 9h}{h} \] Factoring out \( h \) gives: \[ m_{\text{sec}} = 20x + 10h + 9 \] Thus, the final expression for the slope of the secant line in terms of \( x \) and \( h \) is: \[ m_{\text{sec}} = 20x + 10h + 9 \] \( m_{\text {sec }}=\boxed{20x + 10h + 9} \)