Concept 24Essential Question/Topic: I can change a quadraticfrom standard form into vertex form.
24.1 I can change quadratics from standard frominto vertex form.
Convert from standardform to vertex form.
1.“a” is the coefficientof the x2 term.
2.Use the formula tofind h, -b/2a (thesame as the AOSformula!)
3.Substitute the valueyou found for h intothe original equationand solve for k.
Standard Form Vertex Form
Y = ax2 + bx + cy = a(x – h)2 + k
Ex1 Y = 8x2 – 16x + 27
We know a, b, and c. We want a, h, and k.
a = 8
h = -(-16)/2(8)
h = 1
k = (y = 8(x)2 – 16(x) + 27)
k = (y = 8(h)2 – 16(h) + 27)
k = (y = 8(1)2 – 16(1) + 27) = 19
Now, plug all of the values in!
y = a(x – h)2 + k
y = 8(x – 1)2 + 19
Questions/Central Ideas
Notes/Key Details
24.1 I can change quadratics from standard frominto vertex form.
Convert from standardform to vertex form.
1.“a” is the coefficientof the x2 term.
2.Use the formula tofind h, -b/2a (thesame as the AOSformula!)
3.Substitute the valueyou found for h intothe original equationand solve for k.
Standard Form Vertex Form
Y = ax2 + bx + cy = a(x – h)2 + k
Ex2 Y = 5x2 – 40x + 67
We know a, b, and c. We want a, h, and k.
a = 5
h = -(-40)/2(5) = 4
h = 4
k = (y = 5(x)2 – 40(x) + 67)
k = (y = 5(h)2 – 40(h) + 67)
k = (y = 5(4)2 – 40(4) + 67) = -13
Now, plug all of the values in!
y = a(x – h)2 + k
y = 5(x – 4)2 – 13
Questions/Central Ideas
Notes/Key Details
24.1 I can identify the vertex and axis of symmetryfrom vertex form and graph the equation.
Graphing a parabolawhen in vertex form.
1.Find the Axis OfSymmetry. Set theparenthetical equal tozero and solve for x.
2.Find the vertex(lowest or highestpoint of the parabola)by plugging yourAOS in for x.
Vertex Form y = a(x – h)2 + k
Ex1 y = 2(x – 2)2 + 1
(x – 2) = 0
x = 2 Axis of symmetry
y = 2(x – 2)2 + 1
y = 2(2 – 2)2 + 1
y = 2(0)2 + 1
y = 1
Vertex = (2,1)
Questions/Central Ideas
Notes/Key Details
24.1 I can identify the vertex and axis of symmetryfrom vertex form and graph the equation.
3.Find more points toyour parabola bychoosing two x-values on each sideof the vertex.
4.Plug each one of thex-values into yourinitial equation andsolve to get acomplete coordinate.
Vertex = (2,1)
Initial equation: y = 2(x – 2)2 + 1
y = 2(1 – 2)2 + 1
y = 2(-1)2 + 1
y = 2(1) + 1
y = 3
Your coordinate is (1,3). Since it’s a parabola, whenyou plug in a 3 for x, you will also get out 3. Samewith your x-values of 0 and 4, they will have thesame y-values.
Questions/Central Ideas
Notes/Key Details
X
0
1
2
3
4
Y
1
X
0
1
2
3
4
Y
9
3
1
3
9
24.1 I can identify the vertex and axis of symmetryfrom vertex form and graph the equation.
1.Plot the points tograph your parabola!
Questions/Central Ideas
Notes/Key Details
X
0
1
2
3
4
Y
9
3
1
3
9
