Answer Key 10.6

Terrance Berg

Answer Key 10.6

$\phantom{a}$
$\text{intercepts: }\begin{array}[t]{rrl} y&=&0 \\ 0&=&x^2-2x-8 \\ 0&=&(x-4)(x+2) \\ x&=&4,-2 \\ \\ \end{array}$
$\text{vertex: }\begin{array}[t]{l} \left[\dfrac{-b}{2a}, f\left(\dfrac{-b}{2a}\right)\right] \\ \\ (1,-9) \end{array}$
$\text{line of symmetry: }\begin{array}[t]{rll} x&=&\dfrac{-b}{2a} \\ \\ x&=& \dfrac{-(-2)}{2(1)}\Rightarrow \dfrac{2}{2}\text{ or }1 \\ \\ \therefore f(1)&=&1^2-2(1)-8 \\ \phantom{\therefore}f(1)&=&-9 \end{array}$
$\phantom{a}$
$\text{intercepts: }\begin{array}[t]{rrl} 0&=&x^2-2x-3 \\ 0&=&(x-3)(x+1) \\ x&=&3,-1 \end{array}$
$\text{line of symmetry: }\begin{array}[t]{rll} x&=& \dfrac{-(-2)}{2(1)}\Rightarrow \dfrac{2}{2}\text{ or }1 \end{array}$
$\text{vertex: }\begin{array}[t]{rll} f(1)&=&1^2-2(1)-3 \\ f(1)&=&1-2-3 \\ f(1)&=&-4 \\ \\ &&(1,-4) \end{array}$
$\phantom{a}$
$\text{intercepts: }\begin{array}[t]{rrl} 0&=&2x^2-12x+10 \\ 0&=&2(x^2-6x+5) \\ 0&=&2(x-5)(x-1) \\ x&=&5,1 \end{array}$
$\text{line of symmetry: }\begin{array}[t]{rll} x&=&\dfrac{-b}{2a} \\ \\ x&=& \dfrac{-6}{2(1)}\Rightarrow \dfrac{6}{2}\text{ or }3 \end{array}$
$\text{vertex: }\begin{array}[t]{rll} f(3)&=&2(3)^2-12(3)+10 \\ f(3)&=&18-36+10 \\ f(3)&=&-8 \\ \\ &&(3,-8) \end{array}$
$\phantom{a}$
$\text{intercepts: }\begin{array}[t]{rrl} 0&=&2x^2-12x+16 \\ 0&=&2(x^2-6x+8) \\ 0&=&2(x-4)(x-2) \\ x&=&4,2 \end{array}$
$\text{line of symmetry: }\begin{array}[t]{rll} x&=&\dfrac{-b}{2a} \\ \\ x&=& \dfrac{-(-12)}{2(2)}\Rightarrow \dfrac{12}{4}\text{ or }3 \end{array}$
$\text{vertex: }\begin{array}[t]{rll} f(3)&=&2(3)^2-12(3)+16 \\ f(3)&=&18-36+16 \\ f(3)&=&-2 \\ \\ &&(3,-2) \end{array}$
$\phantom{a}$
$\text{intercepts: }\begin{array}[t]{rrl} 0&=&-2x^2+12x-18 \\ 0&=&-2(x^2-6x+9) \\ 0&=&-2(x-3)(x-3) \\ x&=&3 \end{array}$
$\text{line of symmetry: }\begin{array}[t]{rll} x&=&\dfrac{-b}{2a} \\ \\ x&=& \dfrac{-12}{2(-2)}\Rightarrow \dfrac{-12}{-4}\text{ or }3 \end{array}$
$\text{vertex: }\begin{array}[t]{rll} f(3)&=&-2(3)^2-12(3)-18 \\ f(3)&=&-18+36-18 \\ f(3)&=&0 \\ \\ &&(0,3) \end{array}$
$\phantom{a}$
$\text{intercepts: }\begin{array}[t]{rrl} 0&=&-2x^2+12x-10 \\ 0&=&-2(x^2-6x+5) \\ 0&=&-2(x-5)(x-1) \\ x&=&5,1 \end{array}$
$\text{line of symmetry: }\begin{array}[t]{rll} x&=&\dfrac{-b}{2a} \\ \\ x&=& \dfrac{-12}{2(-2)}\Rightarrow \dfrac{-12}{-4}\text{ or }3 \end{array}$
$\text{vertex: }\begin{array}[t]{rll} f(3)&=&-2(3)^2-12(3)-10 \\ f(3)&=&-18+36-10 \\ f(3)&=&8 \\ \\ &&(3,8) \end{array}$
$\phantom{a}$
$\text{intercepts: }\begin{array}[t]{rrl} 0&=&-3x^2+24x-45 \\ 0&=&-3(x^2-8x+15) \\ 0&=&-3(x-3)(x-5) \\ x&=&3,5 \end{array}$
$\text{line of symmetry: }\begin{array}[t]{rll} x&=&\dfrac{-b}{2a} \\ \\ x&=& \dfrac{-24}{2(-3)}\Rightarrow \dfrac{-24}{-6}\text{ or }4 \end{array}$
$\text{vertex: }\begin{array}[t]{rll} f(4)&=&-3(4)^2+24(4)-45 \\ f(4)&=&-48+96-45 \\ f(4)&=&3 \\ \\ &&(4,3) \end{array}$
$\phantom{a}$
$\text{intercepts: }\begin{array}[t]{rrl} 0&=&-2(x^2+2x)+6 \\ 0&=&-2x^2-4x+6 \\ 0&=&-2(x^2+2x-3) \\ 0&=&-2(x+3)(x-1) \\ x&=&-3,1 \end{array}$
$\text{line of symmetry: }\begin{array}[t]{rll} x&=&\dfrac{-b}{2a} \\ \\ x&=& \dfrac{-(-4)}{2(-2)}\Rightarrow \dfrac{4}{-4}\text{ or }-1 \end{array}$
$\text{vertex: }\begin{array}[t]{rll} f(-1)&=&-2(-1)^2-4(-1)+6 \\ f(-1)&=&-2+4+6 \\ f(-1)&=&8 \\ \\ &&(-1,8) \end{array}$
$\text{line of symmetry: }x=\dfrac{-b}{2a}\Rightarrow \dfrac{-(-6)}{2(3)}\Rightarrow \dfrac{6}{6}\text{ or }1$

$x$ $y$

3 4

2 −5

1 −9

0 −5

−1 4
$\text{line of symmetry: }x=\dfrac{-b}{2a}\Rightarrow \dfrac{-(-4)}{2(2)}\Rightarrow \dfrac{4}{4}\text{ or }1$

$x$ $y$

3 3

2 −3

1 −5

0 −3

−1 3
$\text{line of symmetry: }x=\dfrac{-b}{2a}\Rightarrow \dfrac{-4}{2(-1)}\Rightarrow \dfrac{-4}{-2}\text{ or }2$

$x$ $y$

5 −3

4 2

3 5

2 6

1 5

0 2

−1 −3
$\text{line of symmetry: }x=\dfrac{-b}{2a}\Rightarrow \dfrac{-(-6)}{2(-3)}\Rightarrow \dfrac{6}{-6}\text{ or }-1$

$x$ $y$

1 −7

0 2

−1 5

−2 2

−3 −7

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$x$	$y$
3	4
2	−5
1	−9
0	−5
−1	4

$x$	$y$
3	3
2	−3
1	−5
0	−3
−1	3

$x$	$y$
5	−3
4	2
3	5
2	6
1	5
0	2
−1	−3

$x$	$y$
1	−7
0	2
−1	5
−2	2
−3	−7

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