Question 10 Write the equation of the tangent line to the graph of \( y=(2 x+1)^{4} \) at \( x=0 \). The equation must be in slope-intercept form: \( y=m x+b \). Question 11

To find the equation of the tangent line to the graph of \( y = (2x + 1)^4 \) at \( x = 0 \), we need to follow these steps: 1. **Find the value of \( y \) at \( x = 0 \)**. 2. **Calculate the derivative \( y' \) to find the slope of the tangent line at \( x = 0 \)**. 3. **Use the point-slope form of the line to write the equation in slope-intercept form**. ### Step 1: Find \( y \) at \( x = 0 \) Substituting \( x = 0 \) into the equation: \[ y = (2(0) + 1)^4 = (1)^4 = 1 \] So, the point on the graph at \( x = 0 \) is \( (0, 1) \). ### Step 2: Calculate the derivative \( y' \) To find the slope of the tangent line, we need to differentiate \( y = (2x + 1)^4 \). We will use the chain rule for differentiation. Let \( u = 2x + 1 \), then \( y = u^4 \). The derivative \( y' \) is given by: \[ y' = \frac{dy}{du} \cdot \frac{du}{dx} = 4u^3 \cdot 2 = 8u^3 \] Now substituting back \( u = 2x + 1 \): \[ y' = 8(2x + 1)^3 \] Now, we will evaluate \( y' \) at \( x = 0 \): \[ y' = 8(2(0) + 1)^3 = 8(1)^3 = 8 \] So, the slope \( m \) of the tangent line at \( x = 0 \) is \( 8 \). ### Step 3: Write the equation of the tangent line Using the point-slope form of the line, which is \( y - y_1 = m(x - x_1) \), we can substitute \( m = 8 \), \( x_1 = 0 \), and \( y_1 = 1 \): \[ y - 1 = 8(x - 0) \] This simplifies to: \[ y - 1 = 8x \] Adding \( 1 \) to both sides gives us: \[ y = 8x + 1 \] ### Final Answer The equation of the tangent line in slope-intercept form is: \[ y = 8x + 1 \]