Finding the vertex

Finding the vertex

Monday, Jan 4, 2021

Important Dates to Note

• Friday Jan. 8 --> quadratics unit test

• Last day to turn in any late work --> Friday Jan. 22. (this is a hard deadline, I will not accept anything after Jan 22)

Recap

• Parabola --> name of the "U" shape

• x - intercept or Zeros --> where the graph crosses x - axis

• y - intercept --> where the graph crosses y - axis

• Zeros are always written as ordered pair (x, 0) where x is a number

• Y - intercepts are always written as ordered pair (0, y) where y is a number

Finding the vertex

The vertex is the point on the parabola where it changes direction. It's either the highest point or lowest point.

If the equation is in standard form:

 $y=ax^2\ +\ bx\ +\ c$  

To find the x-value use this formula:  $x=\frac{-b}{2a}$  

To find the y value, plug in the x value that you find to the original equation and solve

Open Ended

Find the vertex of

 $y=x^2\ +2x-8$  

Remember:
 $y=ax^2+bx+c,\ \ \ \ \ \ \ \ x=\frac{-b}{2a}$  

Find the vertex of   

 $y=x^2+2x-8$  

•  $x=\frac{-b}{2a}\ \ \ \$  

•  $x=\frac{-2}{2\cdot1}=\frac{-2}{2}=-1$  

•  $y=\left(-1\right)^2+2\left(-1\right)-8$  

•  $y=1-2-8=-9$  

• Vertex is: (-1, -9)

If the equation is in vertex form, the vertex is the values (h, k) - where h is the opposite sign and k is the same

 $y=a\left(x-h\right)^2+k$  

Open Ended

Find the vertex: $y=\left(x-2\right)^2+3$  

Find the vertex of

 $y=\left(x-2\right)^2+3$  

• equation is  $y=a\left(x-h\right)^2+k$  

• vertex is (h, k) where h is 2, and k is 3

• vertex is: (2, 3)

Open Ended

Find the vertex: $y=\left(x+4\right)^2+5$  

find the vertex of

 $y=\left(x+4\right)^2+5$  

• h=-4, k=5

• vertex: (-4, 5)

Finding vertex from a graph

Just look for the highest or lowest point, and write the coordinate.

Open Ended

Find the vertex

Write your answer as an ordered pair

Open Ended

Find the vertex.

Write your answer as an ordered pair

Open Ended

Find the vertex:

Open Ended

Find the vertex: