Finding.zeros.algebraically.vertex.form

Finding Real Zeros of
Quadratic Functions
Day 1 Goals:
•Find the zero(s) of a quadratic function
given in vertex form algebraically, applying
the square root property
•Check the zero(s) with technology.

Cartman was given the following
number sequence:
Choose any number.
Subtract 1 from your number.
Now square your result.
Multiply that number by negative 2.
Finally, add 18.
Cartman’s final number was zero.
Can you figure out his starting
number?

Choose any number.
x
1 x 
 2
x 
1 Multiply that number by negative 2.
 2
2 x 1
Finally, add 18.
 2
2 x 1 18

Choose any number.
Finally, add 18.
 2
2 x 1 18  0
2 2(x 1)  18
2 (x 1)  9
x 1  9
x 1 9
Subtract 18
Divide by -2
Take square root
Add 1

Choose any number.
Finally, add 18.
So, what was Cartman’s starting
number?
x 1 9
x 13
x = 4
…or was it x = -2?

Both x= 4 and x= -2 work!
Verify it with a graph!
f (x)  2(x 1)2 18
( 2, 0)  (4, 0)

So, how do we get two solutions
algebraically?
The Square Root Property
x2  a
If ,
then
x   a

 2
2 x 1 18  0
2 2(x 1)  18
2 (x 1)  9
x 1   9
x 13
x 13 or
x 13
x  4 or
x  2
Square Root Property
We have to include the
± symbol to include
the two possible
values!

Find the zero(s) of the each quadratic
function algebraically. Check your
answer with a graphing calculator.
f (x)  3(x 1)2  48
1 2
2 g(x)  (x 6) 3.125
2 h(x)  4(x  0.45)  2.25
2 k(x)  (3x  5) 1

Finding.zeros.algebraically.vertex.form

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