The vertex of a parabola can be found using the formula based on the quadratic equation in standard form y=ax2+bx+cy=ax^2+bx+cy=ax2+bx+c.
Steps to Find the Vertex:
- Identify coefficients aaa and bbb from the quadratic equation.
- Calculate the x-coordinate of the vertex using the formula:
x=−b2ax=-\frac{b}{2a}x=−2ab
- Substitute this x-coordinate back into the original quadratic equation to find the corresponding y-coordinate.
Thus, the vertex (h,k)(h,k)(h,k) is:
(h,k)=(−b2a, y(−b2a))(h,k)=\left(-\frac{b}{2a},;y\left(-\frac{b}{2a}\right)\right)(h,k)=(−2ab,y(−2ab))
Additional Notes:
- If the quadratic is in vertex form y=a(x−h)2+ky=a(x-h)^2+ky=a(x−h)2+k, the vertex is simply (h,k)(h,k)(h,k).
- For a parabola opening up or down, the vertex represents the minimum or maximum point respectively.
- For a parabola given in factored form y=a(x−p)(x−q)y=a(x-p)(x-q)y=a(x−p)(x−q), the x-coordinate of the vertex is the midpoint of the roots, x=p+q2x=\frac{p+q}{2}x=2p+q, and substituting this value back finds the y-coordinate.
Example:
Given y=2x2−4x+1y=2x^2-4x+1y=2x2−4x+1,
- a=2,b=−4a=2,b=-4a=2,b=−4
- x=−−42×2=1x=-\frac{-4}{2\times 2}=1x=−2×2−4=1
- y=2(1)2−4(1)+1=−1y=2(1)^2-4(1)+1=-1y=2(1)2−4(1)+1=−1
- So vertex is (1,−1)(1,-1)(1,−1).
