2.2.14 Vertex of a Parabola

When approached with a parabola in the form of f(x)=Ax2+Bx+Cf(x) = Ax^2 + Bx + C, the coordinate of the vertex is:

(h,k)=(−B2A,f(−B2A))(h, k) = \left(\frac{-B}{2A}, f\left(\frac{-B}{2A}\right)\right)

Example: Find the y-coordinate of the vertex of the parabola y=3x2−12x+16y = 3x^2 - 12x + 16. x=−(−12)2⋅3=2⇒y=3(2)2−12(2)+16=12−24+16=4x = \frac{-(-12)}{2 \cdot 3} = 2 \Rightarrow y = 3(2)^2 - 12(2) + 16 = 12 - 24 + 16 = 4

It should be noted that if the parabola is in the form x=ay2+by+cx = ay^2 + by + c, then the vertex is: (h,k)=(f(−b2a),−b2a)(h, k) = \left(f\left(\frac{-b}{2a}\right), \frac{-b}{2a}\right)

Problem Set 2.2.14

The vertex of the parabola y = 2x² + 8x - 1 is (h, k), k =
The vertex of y = x² - 2x - 4 is (h, k), k =
If g(x) = 2 - x - x², then the axis of symmetry is x =