Download Algebra I Items to Support Formative Assessment

Algebra I Items to Support Formative Assessment
Unit 3: Quadratic Functions and Modeling
Part II: Algebraic Analysis of Quadratic Functions
A.REI.B.4 Solve quadratic equations in one variable.
a. Use the method of completing the square to transform any quadratic equation in x into
an equation of the form (x – p)2 = q that has the same solutions. Derive the quadratic
formula from this form.
b. Solve quadratic equations by inspection, taking square roots, completing the square, the
quadratic formula and factoring, as appropriate to the initial form of the equation.
Recognize when the quadratic formula gives complex solutions and write them as a+/-bi for
real numbers a and b.
A.REI.B.4b
There are three scenarios you might encounter when solving a quadratic using the quadratic
formula. In the space below, describe each, when they occur algebraically, and what they mean
for the graph of the quadratic.
Solution:
When solving a quadratic equation using the quadratic formula, the determinant can be positive,
negative or zero. If the determinant is zero, then there is one real, rational solution. The graph
of this function would touch the x-axis in one place, at the vertex. If the determinant is positive,
then there are two real solutions. If the determinant is a perfect square, those real solutions will
also be rational. If it’s not a perfect square, then the solutions will be irrational. Either way, the
graph of the function will cross the x-axis in two places. If the determinant is negative, then
there are two, imaginary solutions. The graph of this function will not cross the x-axis at all.
A.REI.B.4ab, F.IF.C.8
Which of the following function(s) is/are equivalent to f (x) = 2x 2 - 4x + 5 ?
f (x) = 2(x + 1)2 + 5
f (x) = 2(x +1)2 - 3
f (x) = 2(x -1)2 + 3
f (x) = 2(x - 2)2 + 5
Solution:
f (x) = 2(x -1)2 + 3
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A.REI.B.4ab
Matthew and Alicia are comparing their class work. They have each solved the same quadratic
function, f (x) = 4x 2 +16x - 8 , for x, but have different answers. Determine if Matthew, Alicia,
both, or neither student is correct. Explain.
Student
Matthew
Work
f (x) = 4x 2 +16x - 8
f (x) = 4(x 2 + 4x + 4) - 8 + 4
f (x) = 4(x + 2)2 - 4
0 = 4(x + 2)2 - 4
1 = (x + 2)2
x = -2 ± 1
x = -3,-1
Alicia
f (x) = 4x 2 +16x - 8
f (x) = (4x 2 +16x + 64) - 8 - 64
f (x) = (4x + 8)2 - 72
0 = (4x + 8)2 - 72
72 = (4x + 8)2
x=
-8 ± 72
4
x = -2 ±
3 2
2
Solution:
Neither Matthew nor Alicia are correct. Matthew made an error when he added four to the end of
the equation. He should have subtracted 16. Therefore his remaining work and solution are
wrong. Alicia did not factor out the lead coefficient, then added & subtracted the wrong value,
and therefore is also incorrect. She should have factored out the 4 from the ax2 and bx terms
before calculating the value to complete the square.
The correct answer is:
f (x) = 4x 2 +16x - 8
f (x) = 4(x 2 + 4x + 4) - 8 - 16
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product under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.
f (x) = 4(x + 2)2 - 24
0 = 4(x + 2)2 - 24
6 = (x + 2)2
x = -2 ± 6
A.REI.B.4ab, F.IF.C.8
Write the function f (x) = x 2 + 6x - 4 in vertex form. Then find the zeros.
Solution:
f (x) = (x 2 + 6x + 9) - 4 - 9
f (x) = (x + 3)2 - 13
0 = (x + 3)2 - 13
x = -3 ± 13
A.REI.B.4ab
Solve the following equations. For each equation, use a different strategy to find the solution.
a. 3x 2 +14x = 5
b. x 2 + 4x + 7 = 0
c. x 2 = 5 - 4x
For each of the above, explain why you chose each method to solve.
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product under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.
Possible Solution:
3x 2 +14 x - 5 = 0
-14 ± 196 - 4(3)(-5)
6
-14 ± 196 + 60
x=
6
-14 ± 16
x=
6
1
x = -5,
3
x=
x2 + 4x - 7 = 0
x2 = 5 - 4x
(x 2 + 4 x + 4) = 7 + 4 x 2 + 4 x - 5 = 0
(x + 2) 2 = 11
(x + 5)(x -1) = 0
x = -2 ± 11
x = -5,1
Explanation – For the first equation, the lead coefficient of 3 makes it difficult to solve by
factoring, and completing the square would require that we factor out the three. That’s prone to
error, so the quadratic formula helps avoid that situation. The third problem is easily factored
into two binomials by inspection. That leaves the second equation to be solved by completing
the square. The irrational solution that resulted confirmed that it could not be factored over the
integers.
Howard County Public Schools Office of Secondary Mathematics Curricular Projects has licensed this
product under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.