What is the equation of the line normal to #f(x)= x^3+4x^2 # at #x=1#?

Question

Emma Aldridge · Accepted Answer

#y-5=-1/11(x-1)#

We'll be using the point gradient formula since we already have a point, x=1, which means y=5; #(1,5)# no we need a gradient.

In order to find the gradient at x=1. we need to differentiate the equation. The differentiated equation is #3x^2+8x#. We then substitute in x=1, and we have the gradient of 11.

However, this gradient is the tangents gradient, not the normal's. In order to find the normal's gradient, we know that the gradient of the tangent times the gradient of the normal equals -1. Therefore, the gradient of the normal is #-1/11#

We the substitute the numbers into the formula to get a equation

Cameron Alexander · Answer

undefined The equation of the line normal to f(x) = x^3 + 4x^2 at x = 1 is y = -10x + 14.

Emily Ammerman · Answer

undefined To find the equation of the line normal to the function $ f(x) = x^3 + 4x^2 $ at $ x = 1 $, we first need to find the slope of the tangent line to the function at $ x = 1 $, and then find the negative reciprocal of this slope to get the slope of the normal line.

1. Find the derivative of the function $ f(x) = x^3 + 4x^2 $ using the power rule:
$$ f'(x) = 3x^2 + 8x $$

2. Evaluate the derivative at $ x = 1 $ to find the slope of the tangent line:
$$ f'(1) = 3(1)^2 + 8(1) = 3 + 8 = 11 $$

3. Since the normal line is perpendicular to the tangent line, its slope will be the negative reciprocal of the slope of the tangent line:
$$ m_{\text{normal}} = -\frac{1}{11} $$

Now, we have the slope of the normal line and a point on the line, which is $ (1, f(1)) $.

4. Substitute $ x = 1 $ into the original function $ f(x) = x^3 + 4x^2 $ to find the corresponding $ y $-coordinate:
$$ f(1) = (1)^3 + 4(1)^2 = 1 + 4 = 5 $$

So, the point on the normal line is $ (1, 5) $.

5. Now, we use the point-slope form of the equation of a line, where $ (x_1, y_1) $ is the point on the line and $ m $ is the slope:
$$ y - y_1 = m(x - x_1) $$

Substituting $ (1, 5) $ for $ (x_1, y_1) $ and $ -\frac{1}{11} $ for $ m $:
$$ y - 5 = -\frac{1}{11}(x - 1) $$

6. Simplify the equation:
$$ y - 5 = -\frac{1}{11}x + \frac{1}{11} $$
$$ y = -\frac{1}{11}x + \frac{56}{11} $$

So, the equation of the line normal to $ f(x) = x^3 + 4x^2 $ at $ x = 1 $ is $ y = -\frac{1}{11}x + \frac{56}{11} $.