MTH 234 Groupwork 3

MTH 234 Groupwork 3 - 2.2, 2.3
(Your goal today is for you to prove the quadratic formula)
Understanding Parabolas as Translations
Parabola: f (x) = ax2 + bx + c
1. Use the information from lecture about vertical and horizontal translations to graph the following parabolas on the axes provided below.
Label the coordinates of vertex of each parabola on the graph.
(a) y = x2
(b) y = (x − 5)2 + 3
(c) y = (x + 5)2 − 3
2. Explain (using translations and the examples above) why when a
quadratic equation is written in y = a(x − h)2 + k form, the vertex of the parabola can be read as (h,k). We will call this form of a
quadratic function vertex form. (Don’t worry about what the letter
a is doing there yet)
1
3. A quadratic function can be put into vertex form by a process called
completing the square. Watch the steps below that put the given
quadratic function in vertex form.
Notice what a perfect square looks like multiplied out
b 2
x+
= x+
2
b 2
x−
= x−
2
2
b
b
2
x+
= x + bx +
2
2
2
b
b
b
2
x−
= x − bx +
2
2
2
b
2
(Start with a quadratic equation like below)
y = 2x2 − 12x + 23
(Subtract the constant from both sides)
y − 23 = 2x2 − 12x
(Factor out the leading coefficient from the right-hand side)
y − 23 = 2(x2 − 6x)
(Complete the square inside the parenthesis - this means
adding the same number on the left. Watch out - this is the
hardest part!)
y − 23 + 2(9) = 2(x2 − 6x + 9)
(Write the right-hand side as a perfect square)
y − 5 = 2(x − 3)2
(Move the new constant term back to the right side)
y = 2(x − 3)2 + 5
Now use the vertex formula from lecture to check that the
vertex of y = 2x2 − 12x + 23 is (3,5)
4. Follow the steps above to put a general quadratic equation in vertex
form.
(Start with the quadratic equation below)
y = ax2 + bx + c
(Subtract the constant from both sides)
(Factor out the leading coefficient from the right-hand side)
2
(Complete the square inside the parenthesis - this means adding the
same number on the left)
(Write the right-hand side as a perfect square)
(Use common denominators to simplify the left-hand side and
combine constants)
(Move the new constant term back to the right side)
b 2 b2 − 4ac
y =a x+
−
2a
4a
5. We will use what we worked out above now to find all x-intercepts
(roots) of a quadratic equation.
(Start with the quadratic equation below)
y = ax2 + bx + c
(Write what you would get after you rearrange it into vertex form)
b 2 b2 − 4ac
y =a x+
−
2a
4a
(X-intercepts happen when y=0. Set y equal to zero below)
(Start solving for x by moving the constant to the left-hand side)
(Divide both sides by a)
(Take the square root of both sides)
r
±
b2 − 4ac
=
4a2
3
(Simplify the radical on the left-hand side
(Subtract -b/2a from both sides)
(Use common denominators to add fractions on the left-hand side)
4

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