For the function \( y=2 x^{2}+7 x+3 \), at the point \( x=8 \), find the following, (a).the.slage of the tangent to the curve Enter an exact number (b) the instantaneous rate of change of the function

To find the slope of the tangent to the curve for the function \( y = 2x^2 + 7x + 3 \) at the point \( x = 8 \), we need to first compute the derivative of the function: 1. The derivative \( y' = \frac{dy}{dx} = 4x + 7 \). Now, substituting \( x = 8 \) into the derivative: \[ y'(8) = 4(8) + 7 = 32 + 7 = 39. \] Thus, the slope of the tangent to the curve at \( x = 8 \) is \( \text{39} \). For the second part, the instantaneous rate of change of the function at \( x = 8 \) is the same as the slope of the tangent, which is also \( \text{39} \).