Question

the slope of the tangent line to the parabola $y = 4x^2 - 3x + 6$ at the point where $x = 0$ is:

the equation of this tangent line can be written in the form $y = mx + b$ where $m$ is:
-3
and where $b$ is: 6
question help: video

Explanation:

Step1: Find the derivative of the function

The function is \( y = 4x^2 - 3x + 6 \). The derivative of \( y \) with respect to \( x \), using the power rule (\( \frac{d}{dx}(x^n)=nx^{n - 1} \)), is \( y'=8x-3 \).

Step2: Evaluate the derivative at \( x = 0 \)

To find the slope of the tangent line at \( x = 0 \), we substitute \( x = 0 \) into the derivative. So \( y'(0)=8(0)-3=-3 \).

Step3: Find the point on the parabola at \( x = 0 \)

Substitute \( x = 0 \) into the original function \( y = 4(0)^2-3(0)+6 = 6 \). So the point is \( (0,6) \).

Step4: Use the point - slope form to find the tangent line equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(0,6) \) and \( m=-3 \). Substituting these values, we get \( y - 6=-3(x - 0) \), which simplifies to \( y=-3x + 6 \). In the form \( y=mx + b \), \( m=-3 \) and \( b = 6 \). For the slope of the tangent line (first blank), from Step 2, the slope is \(-3\).

Answer:

The slope of the tangent line is \(-3\), \( m=-3 \), \( b = 6 \)