What is the equation of the tangent line to f(x) = x² - 1 at the point (-2, 3)?
2007-09-17 15:52:17 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
First, you need to find the derivative of the original equation. It is y'=2x
Next, plug in your point into the new equation of the tangent line to find B in y=mx+b
So, (3)=2(-2)+b
solve for b
so 3=-4+b
b=7
your final tangent line equation at the point (-2,3) is y'=2x+7
2007-09-17 16:03:22 · answer #1 · answered by Anonymous · 0⤊ 1⤋
slope of tangent to f(x) = x² - 1 is derivative d(f(x))/dx = 2x, so at (-2,3), slope is -4.
So y - 3 = -4(x + 2)
y = -4x - 8 + 3
y = -4x - 5.
2007-09-17 23:03:41 · answer #2 · answered by Philo 7 · 1⤊ 0⤋
i dont no about tangents, so i dont wanna say any wrong answer. but i think u can use a graphing calculator to solve this one. im 99 percent sure.
2007-09-17 22:59:47 · answer #3 · answered by Harris 6 · 0⤊ 1⤋
The person right above me is correct.
2007-09-17 23:05:38 · answer #4 · answered by vendetta4hire 3 · 0⤊ 0⤋