Question

guided practice
the quadratic function r(x) and q(x) are given below. the function with the smaller minimum value is:

x	r(x)
-4	-12
-3	-15
-2	-16
-1	-15
0	-12
1	7

$q(x) = x^2 + 2x - 8$
a. $q(x)$ and the value of $-9$
b. $q(x)$ and the value of $-1$
c. $r(x)$ and the value of $-16$
d. $r(x)$ and the value of $-2$

purposeful practice

1. which of the quadratic functions below has the smallest minimum value?

a. $h(x) = x^2 + 2x - 6$
b. image of a parabola on a grid
c. $h(x) = (x + 5)(x + 2)$
d. | x | f(x) |

-1	-2
0	-5
1	-6
2	-5
3	-2

Explanation:

Step1: Analyze function a

For \( h(x) = x^2 + 2x - 6 \), complete the square:
\( h(x) = (x + 1)^2 - 7 \).
The minimum value (vertex y - coordinate) is \(-7\).

Step2: Analyze function b (graph)

The graph is a parabola opening up. Visually, the vertex (minimum point) appears to have a y - value greater than \(-7\) (e.g., from the grid, it seems above \(-7\)).

Step3: Analyze function c

Expand \( h(x)=(x + 5)(x + 2)=x^2+7x + 10 \).
Complete the square: \( x^2+7x + 10=(x+\frac{7}{2})^2-\frac{49}{4}+10=(x + 3.5)^2-\frac{9}{4}=-2.25 \). Wait, correction: \( (x+\frac{7}{2})^2=\ x^2 + 7x+\frac{49}{4} \), so \( x^2+7x + 10=(x+\frac{7}{2})^2-\frac{49}{4}+10=(x + 3.5)^2-\frac{49 - 40}{4}=(x + 3.5)^2-\frac{9}{4}=-2.25 \)? No, wait \( 10=\frac{40}{4} \), so \( -\frac{49}{4}+\frac{40}{4}=-\frac{9}{4}=-2.25 \)? Wait, no, that's wrong. Wait, \( (x + 5)(x + 2)=x^2+7x + 10 \). The vertex x - coordinate is \( x=-\frac{b}{2a}=-\frac{7}{2}=-3.5 \). Then \( h(-3.5)=(-3.5 + 5)(-3.5 + 2)=(1.5)(-1.5)=-2.25 \). So minimum value is \(-2.25\)? Wait, no, earlier mistake. Wait, \( a = 1>0 \), so parabola opens up, vertex is minimum. Wait, but when we expand \( (x + 5)(x + 2)=x^2+7x + 10 \), the vertex y - value is \( h(-\frac{7}{2})=(-\frac{7}{2}+5)(-\frac{7}{2}+2)=(\frac{3}{2})(-\frac{3}{2})=-\frac{9}{4}=-2.25 \). Wait, but earlier for function a, the minimum was \(-7\), which is less than \(-2.25\). Wait, I made a mistake in function c's expansion? Wait no, \( (x + 5)(x + 2)=x^2+7x + 10 \), correct. Then vertex at \( x=-\frac{7}{2}=-3.5 \), \( y=(-3.5)^2+7*(-3.5)+10=12.25-24.5 + 10=-2.25 \).

Step4: Analyze function d (table)

The table for \( f(x) \) has values: at \( x=-1, f(x)=-2 \); \( x = 0, f(x)=-5 \); \( x = 1, f(x)=-6 \); \( x = 2, f(x)=-5 \); \( x = 3, f(x)=-2 \). The vertex (minimum) is at \( x = 1 \), \( f(1)=-6 \), which is greater than \(-7\).

Wait, correction for function a: \( h(x)=x^2 + 2x-6=(x + 1)^2-7 \), so minimum is \(-7\).
Function c: \( h(x)=x^2+7x + 10=(x+\frac{7}{2})^2-\frac{49}{4}+10=(x + 3.5)^2-\frac{9}{4}=-2.25 \) (minimum).
Function d: minimum at \( x = 1 \), \( f(1)=-6 \).
Function b: from the graph, the vertex y - value is, looking at the graph, the parabola has its vertex above \( y=-7 \) (since the grid lines, the lowest point seems to be at a y - value higher than \(-7\), like maybe \(-5\) or so).

So comparing the minimum values:

• Function a: \(-7\)
• Function b: \(>-7\)
• Function c: \(-2.25\) (which is \(-2.25>-7\))
• Function d: \(-6\) (which is \(-6>-7\))

Answer:

a. \( h(x)=x^2 + 2x - 6 \)