Question

Find the equation of the normal to the curve y=x²+4x-3 at point(1,2)

273

likes1367 views

Maude

4.7

67 Answers

First, find the derivative of the curve to get the slope of the tangent line.
y = x² + 4x - 3
y' = 2x + 4
Evaluate the derivative at the point (1, 2) to find the slope of the tangent line at that point.
y'(1) = 2(1) + 4 = 2 + 4 = 6
So, the slope of the tangent line at (1, 2) is 6.
The slope of the normal line will be the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is -1/6.
Now that you have the slope of the normal line and the point (1, 2), you can use the point-slope form of a line to find the equation of the normal line:
y - y₁ = m(x - x₁)
Where: (x₁, y₁) = (1, 2) m = -1/6 (slope of the normal line)
Now, plug in the values:
y - 2 = (-1/6)(x - 1)
Multiply both sides by 6 to eliminate the fraction:
6(y - 2) = -1(x - 1)
6y - 12 = -x + 1
Now, rearrange the equation to get it in standard form:
x + 6y = 13

Frequently asked questions (FAQs)

Math Question: In triangle ABC, the angle bisector of angle A intersects side BC at point D. If BD = 4cm and CD = 6cm, what is the length of side AC?

+

What is the result of adding vector A (3i - 4j) to vector B (-2i + 6j) and subtracting vector C (7i + 2j)?

+

Math question: What is the maximum number of turning points a cubic function, f(x) = x^3, can have within its domain?

+

New questions in Mathematics