Question

question 19 (1 point) determine the y - intercept of a line with a slope of 1 that is a tangent to the curve y = x^2+3x - 3. a) 4 b) - 2 c) - 3 d) - 4

Explanation:

Step1: Differentiate the curve function

The derivative of $y = x^{2}+3x - 3$ using the power - rule $(x^n)'=nx^{n - 1}$ is $y'=2x + 3$.

Step2: Find the x - value of the tangent point

Since the slope of the tangent line is 1, set $y'=1$. So, $2x+3 = 1$. Solving for $x$:
\[

$$\begin{align*} 2x&=1 - 3\\ 2x&=-2\\ x&=-1 \end{align*}$$

\]

Step3: Find the y - value of the tangent point

Substitute $x=-1$ into the original curve function $y = x^{2}+3x - 3$. Then $y=(-1)^{2}+3\times(-1)-3=1 - 3 - 3=-5$.

Step4: Find the equation of the tangent line

The equation of a line in slope - intercept form is $y=mx + b$, where $m = 1$ (the slope) and the line passes through the point $(-1,-5)$. Substitute $x=-1$, $y=-5$ and $m = 1$ into $y=mx + b$:
\[

$$\begin{align*} -5&=1\times(-1)+b\\ -5&=-1 + b\\ b&=-4 \end{align*}$$

\]
The y - intercept of the tangent line is $-4$.

Answer:

d) - 4