Final answer:
To find the equation of the line tangent to the curve y = 3x² + 12x at x = -1, we take the derivative of the curve and evaluate it at x = -1. The slope is 6, so we can use a point-slope form to find the equation of the tangent line. The x-intercept is (3/2, 0) and the y-intercept is (0, -9).
Step-by-step explanation:
To find the equation of the line tangent to the curve y = 3x² + 12x at x = -1, we need to find the derivative of the curve with respect to x and evaluate it at x = -1. The derivative of y = 3x² + 12x is dy/dx = 6x + 12. Plugging in x = -1, we get dy/dx = 6(-1) + 12 = 6.
So, the slope of the tangent line is 6. Now we can use the point-slope form of a line to find the equation. The point-slope form is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope. We can choose (-1, f(-1)) as our point, where f(x) is the given curve.
Plugging in x = -1, we get y = 3(-1)² + 12(-1) = -3 - 12 = -15. So, our point is (-1, -15). Now we can substitute the values into the point-slope form to find the equation: y - (-15) = 6(x - (-1)), which simplifies to y + 15 = 6(x + 1).
To find the x-intercept of the tangent line, we set y = 0 and solve for x. 0 + 15 = 6(x + 1), 15 = 6x + 6, 9 = 6x, x = 9/6 = 3/2. So, the x-intercept is (3/2, 0).
To find the y-intercept, we set x = 0 and solve for y. y + 15 = 6(0 + 1), y + 15 = 6, y = 6 - 15 = -9. So, the y-intercept is (0, -9).