Question

question 34 (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: Find the derivative of the curve

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

Step2: Set the derivative equal to the slope of the tangent line

Since the slope of the tangent line is 1, we set $2x+3 = 1$.
$2x=1 - 3=-2$, so $x=-1$.

Step3: Find the y - coordinate of the point of tangency

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

Step4: Use the point - slope form of a line to find the y - intercept

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-1,-5)$ and $m = 1$.
$y+5=1\times(x + 1)$, which simplifies to $y=x+1 - 5=x - 4$.
The y - intercept is the value of $y$ when $x = 0$. So the y - intercept is $-4$.

Answer:

d) -4