# How do I find the tangent line of y=x^2 at its vertex?

Sep 12, 2014

Let's start by finding the vertex. Then we find the derivative of the function using the x value of the vertex. Lastly, we use the slope intercept formula and the value found for the derivative to create the equation for the tangent line.

We can use the expressions $\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$ to find the vertex.

$y = {x}^{2}$

$a = 1 , b = 0 , c = 0$

$x = - \frac{0}{2 \left(1\right)} = \frac{0}{2} = 0$

$y = f \left(0\right) = {0}^{2} = 0$

Vertex: $\left(0 , 0\right)$

Derivative:

$y = f \left(x\right) = {x}^{2}$

$y ' = f ' \left(x\right) = 2 x$

Substitute in the x value of the vertex, which is 0

$f ' \left(0\right) = 2 \left(0\right) = 0$ This is the slope aka derivative

Find the equation for the Tangent line:

$y = m x + b$, slope intercept formula

Substitute in $x$ and $y$ from the Vertex

$0 = m \left(0\right) + b$

Substitute in the slope/derivative

$0 = \left(0\right) \left(0\right) + b$

Solve for b, which is also known as the y-intercept

$0 = b$

Now that we know the value of $b$ we can create the general form of the tangent line equation by just substituting in the slope/derivative and the b (y-intercept) value.

$y = \left(0\right) x + 0$, Next we simplify

Solution:

$y = 0$ , the equation of the tangent line