Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
2. Quadratic Functions
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Exercise 37 Page 558

When a multiple choice questions asks for lowest, substitute the answers into the question to starting with the lowest coordinate on the list.

C

Practice makes perfect
To find the lowest point of a quadratic, we need to find the vertex. Let's start by finding the axis of symmetry, which will give us the x-value of the vertex. x=- b/2 aIn our function y=x^2-6x+9 we can substitute a= 1 and b= -6 into the formula for the axis of symmetry.
x=- b/2 a
x=-( -6)/2( 1)
â–Ľ
Calculate x
x=6/2
x=3
The axis of symmetry gives us the x-coordinate for vertex, x=3. Although there is only one answer choice with an x-coordinate of 3 we will continue with the direct solution. We can substitute x=3 into our function to find the y-value of the vertex.
y=x^2-6x+9
y=( 3)^2-6( 3)+9
â–Ľ
Calculate y
y=9-6(3)+9
y=9-18+9
y=0
From this, we can conclude that our minimum point of the ramp is (3,0), answer choice C.

Alternative Solution

Working Backwards

One effective way to manage multiple choice questions is to use your answer choices, rather than solving the question directly. Since the question is is asking about the lowest point, we can start with answer choice B ( 36, -3) which has the least y-value. In this case x= 36 and y= -3. ccc y&=& x^2&-&6 x&+&9 ↕ & &↕ && ↕ && ↕ -3& = & 36^2&-&6( 36)& +&9 & FALSE Since 36^2-6( 36) +9 ≠ -3, we can try the next lowest answer choice, C ( 3, 0). ccc y&=& x^2&-&6 x&+&9 ↕ & &↕ && ↕ && ↕ 0& = & 3^2&-&6( 3)& +&9 & TRUE Here we see that (3,0) is a point on the parabola, and because it is the lowest point of the answer choices that works, C is the correct answer.