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Algebra 2
Vertex Form
Vertex Form: Notes and Examples
When a quadratic function is in vertex form, f(x) = a(x – h)2 + k,
the vertex is at (h, k).
Notice that vertex form looks very similar to point-slope form for linear functions.
As with standard form, you can tell the graph shape from the a value:  if a is positive,  if a is
negative.
Examples:
f(x) = 3(x – 2)2 + 5
f(x) = 3(x + 2)2 + 5
f(x) = 3(x – 2)2 – 5
f(x) = 3(x + 2)2 – 5
 shape; vertex at (2, 5)
 shape; vertex at (–2, 5)
 shape; vertex at (2, –5)
 shape; vertex at (–2, –5)
Sketching graphs
You can use knowledge of the vertex and direction of a quadratic to make a rough sketch of the
parabola graph.
Example: Sketch the graph of f(x) = –2(x + 3)2 + 4.
The vertex is (–3, 4).
The shape is  because the –2 is negative.
Here’s a sketch:
Equation solving involving vertex form
When you have to solve an equation where one side is a quadratic in vertex form, the solving can
be done through steps applied to both sides of the equation. A key step is square-rooting both
sides of the equation; when you take this step, you have to put in a ±.
Example: Solve 2(x – 5)2 + 3 = 35.
Start with:
Subtract 3:
Divide by 2:
Square root:
Split into two equations:
Finish solving:
2(x – 5)2 + 3 = 35
2(x – 5)2
= 32
(x – 5)2
= 16
(x – 5)
= ±4
x–5=4
x
=9
important: notice the ±
x – 5 = –4
x
= 1
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Algebra 2
Vertex Form
Homework Problems:
1. Identify the vertex for each of these quadratic functions.
a. f(x) = –2(x – 1)2 + 6
b. f(x) =
c. f(x) = (x + 4)2 + 2
d. f(x) = –(x – 1)2 – 7
e. f(x) = 5(x – 3)2
f.
2
3
(x + 5)2 – 3
f(x) = 4 x2 + 2
2. Below are several quadratic functions written in vertex form. Without using your calculator,
sketch a graph for each function, and label the vertex with its coordinates. (See the previous
page for an example.)
a. f(x) = 2(x – 1)2 + 3
b. f(x) = –3(x + 2)2 + 4
c. f(x) = (x – 4)2 – 5
d. f(x) = – 12 (x + 3)2 – 1
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e. f(x) = –2(x + 5)2
Algebra 2
Vertex Form
f. f(x) = 4x2 + 3
3. Check your answers to problem 3 by graphing each function on your calculator. (Your graph
might not look exactly like the calculator screen, but you should have the correct vertex and
the correct choice of  vs. .)
4. Solve these equations, following the model of the example on page 1. You’ll need to use a
square-root step; don’t forget the ± that has to be put in at that step.
a. Solve (x + 8)2 + 5 = 30
b. Solve 14 = 2(x – 5)2 – 4
c. Solve (x – 2)2 = 16
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d. Solve (x – 2)2 = 13
e. Solve 4 = 2(x + 3)2
f. Solve –12 = –3(x – 1)2
g. Solve – (x + 2)2 + 10 = –26
h. Solve –30 = –5(x – 4)2
i. Solve 24 = 8x2
Algebra 2
Vertex Form
Hint: 13 not a whole number, but get it from calculator.
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Algebra 2
Vertex Form
Application Problem
5. A rocket is launched into the air. Its height after x seconds is given by
h(x) = –16(x – 4)2 + 256
a. How high is the rocket, 1 second after it’s launched?
b. What is the greatest height reached by this rocket, and at what time does it happen?
c. At what times x is the rocket’s height 240 feet?
Hint: Solve the equation 240 = –16(x – 4)2 + 256
d. At what times x is the rocket’s height 192 feet?
e. At what times x is the rocket’s height 100 feet?
Hint: This time you’ll end up needing a non-whole-number square root from your calculator.