Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Mid-Chapter Quiz
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Exercise 25 Page 224

Write down both expressions and look at their structure. What operations can you do to make one look more like the other?

See solution.

Practice makes perfect

We can write any quadratic equation using the standard form of a quadratic function or the vertex form. Standard Form 1cm Vertex form ax^2+bx +c 1.35cma(x-h)^2+k In these equations, a, b, c, h, and k are real numbers, and a≠0. We will discuss how to take one form to the other.

Going From Vertex Form to Standard Form

Let's start by discussing how to go from the vertex form to the standard form first. For this we need to expand the square binomial and distribute the parameter a. that would be enough to obtain an expression in standard form.
a(x-h)^2+k
a(x^2-2hx+h^2) + k
ax^2-2ahx+ah^2 + k
Notice that the expression we just obtained is in standard form, as required. ax^2+bx+c ax^2-2ahx+ ah^2 + k We can see a worked example below.
2(x-3)^2+1
2(x^2-6x+9)+1
2x^2-12x+18 + 1
2x^2-12x+19
The expression 2(x-3)^2+1 in vertex form is equivalent to the expression 2x^2-12x+19, which is in standard form.

Going From Standard Form to Vertex Form

To go from standard form to vertex form, the process is a bit more elaborate. We can start by comparing the corresponding coefficients from the results we found before. This will let us know their equivalences. a = a b = -2ah c=ah^2+k We can already see that the a parameter is the same. Now, we can solve for h from one of the equalities shown above. For simplicity, we will do it from b = - 2ah.
b = -2ah
â–Ľ
Solve for h
b/-2a = h
-b/2a =h
h = - b/2a
Now we can use this result in the other equality and solve for k.
c=ah^2+k
c = a (- b/2a)^2 + k
â–Ľ
Solve for k
c = a b^2/4a^2 + k
c = ab^2/4a^2 + k
c = b^2/4a + k
c -b^2/4a = k
k = c -b^2/4a
We can also rewrite the result as a single fraction by multiplying and dividing c by 4a, getting the result k = 4ac - b^24a. With the results obtained, we can go from standard form to vertex form. Standard Form 1cm Vertex form ax^2+bx +c 1.35cma(x-h)^2+k Equivalence between parameters a=a h = - b/2a k = 4ac - b^2/4a ax^2+bx +c → a(x-h)^2+k 0.5cm a(x - (-b/2a))^2 + 4ac - b^2/4a Let's try an example. We are going to rewrite the same expression we obtained above, 4x^2-12x+19, in vertex form. For this, we start by identifying the parameters from the standard form. 2x^2-12x+ 19 ax^2+bx+c a = 2 b = -12 c=19 Now we follow the formula we found before.
a(x - (-b/2a))^2 + 4ac - b^2/4a
2(x - (--12/2(2)))^2 + 4(2)(19) - (-12)^2/4(2)
â–Ľ
Simplify
2(x - (--12/2(2)))^2 + 4(2)(19) - 144/4(2)
2(x - (--12/4))^2 + 152 - 144/8
2(x - (-12/-4))^2 + 152 - 144/8
2(x - 3)^2 + (19 -18)
2(x - 3)^2 + 1
The corresponding vertex form for the expression is 2(x - 3)^2 + 1. These results agree with our previous results, when we rewrote the same expression from vertex form to standard form.