The function is
f ( x ) = { 24 x < 0 a x 2 + b x + c x ≥ 0 f(x)= \begin{cases}
24 & x<0 \\
ax^2+bx+c &x\geq0
\end{cases} f ( x ) = { 24 a x 2 + b x + c x < 0 x ≥ 0 f ( 0 ) = 24 : 24 = c f(0)=24: 24=c f ( 0 ) = 24 : 24 = c
f ( 5 ) = − 1 : − 1 = a ( 5 ) 2 + b ( 5 ) + 24 f(5)=-1: -1=a(5)^2+b(5)+24 f ( 5 ) = − 1 : − 1 = a ( 5 ) 2 + b ( 5 ) + 24
f ( 10 ) = 26 : 26 = a ( 10 ) 2 + b ( 10 ) + 24 f(10)=26: 26=a(10)^2+b(10)+24 f ( 10 ) = 26 : 26 = a ( 10 ) 2 + b ( 10 ) + 24
25 a + 5 b = − 25 100 a + 10 b = 2 \begin{matrix}
25a+5b=-25 \\
100a+10b=2
\end{matrix} 25 a + 5 b = − 25 100 a + 10 b = 2
b = − 5 a − 5 25 a = 26 \begin{matrix}
b=-5a-5 \\
25a=26
\end{matrix} b = − 5 a − 5 25 a = 26
a = 1.04 b = − 10.2 \begin{matrix}
a=1.04 \\
b=-10.2
\end{matrix} a = 1.04 b = − 10.2
f ( x ) = { 24 x < 0 1.04 x 2 − 10.2 x + c x ≥ 0 f(x)= \begin{cases}
24 & x<0 \\
1.04x^2-10.2x+c &x\geq0
\end{cases} f ( x ) = { 24 1.04 x 2 − 10.2 x + c x < 0 x ≥ 0 x v e r t e x = − b 2 a = − − 10.2 2 ( 1.04 ) = 255 52 x_{vertex}=-\dfrac{b}{2a}=-\dfrac{-10.2}{2(1.04)}=\dfrac{255}{52} x v er t e x = − 2 a b = − 2 ( 1.04 ) − 10.2 = 52 255
y v e r t e x = 1.04 ( 255 52 ) 2 − 10.2 ( 255 52 ) + 24 = − 105 104 y_{vertex}=1.04(\dfrac{255}{52})^2-10.2(\dfrac{255}{52})+24=-\dfrac{105}{104} y v er t e x = 1.04 ( 52 255 ) 2 − 10.2 ( 52 255 ) + 24 = − 104 105
f ( 0 ) = 24 f(0)=24 f ( 0 ) = 24
y = 0 : 1.04 x 2 − 10.2 x + 24 = 0 y=0: 1.04x^2-10.2x+24=0 y = 0 : 1.04 x 2 − 10.2 x + 24 = 0
1.04 x 2 − 10.2 x + 24 = 0 1.04x^2-10.2x+24=0 1.04 x 2 − 10.2 x + 24 = 0
26 x 2 − 255 x + 600 = 0 26x^2-255x+600=0 26 x 2 − 255 x + 600 = 0
x = 255 ± ( − 255 ) 2 − 4 ( 26 ) ( 600 ) 2 ( 26 ) = 255 ± 5 105 52 x=\dfrac{255\pm\sqrt{(-255)^2-4(26)(600)}}{2(26)}=\dfrac{255\pm5\sqrt{105}}{52} x = 2 ( 26 ) 255 ± ( − 255 ) 2 − 4 ( 26 ) ( 600 ) = 52 255 ± 5 105
x 1 = 255 − 5 105 52 , x 2 = 255 + 5 105 52 x_1=\dfrac{255-5\sqrt{105}}{52}, x_2=\dfrac{255+5\sqrt{105}}{52} x 1 = 52 255 − 5 105 , x 2 = 52 255 + 5 105
f ( x ) = 3 = 1.04 x 2 − 10.2 x + 24 f(x)=3=1.04x^2-10.2x+24 f ( x ) = 3 = 1.04 x 2 − 10.2 x + 24
1.04 x 2 − 10.2 x + 21 = 0 1.04x^2-10.2x+21=0 1.04 x 2 − 10.2 x + 21 = 0
26 x 2 − 255 x + 525 = 0 26x^2-255x+525=0 26 x 2 − 255 x + 525 = 0
x = 255 ± ( − 255 ) 2 − 4 ( 26 ) ( 525 ) 2 ( 26 ) = 255 ± 5 417 52 x=\dfrac{255\pm\sqrt{(-255)^2-4(26)(525)}}{2(26)}=\dfrac{255\pm5\sqrt{417}}{52} x = 2 ( 26 ) 255 ± ( − 255 ) 2 − 4 ( 26 ) ( 525 ) = 52 255 ± 5 417
x 1 = 255 − 5 417 52 , x 2 = 255 + 5 417 52 x_1=\dfrac{255-5\sqrt{417}}{52}, x_2=\dfrac{255+5\sqrt{417}}{52} x 1 = 52 255 − 5 417 , x 2 = 52 255 + 5 417
(a) Domain: ( − ∞ , ∞ ) (-\infin,\infin) ( − ∞ , ∞ )
(b) Range: ( − 105 104 , ∞ ) ( -\dfrac{105}{104}, \infin) ( − 104 105 , ∞ )
(c) the x-intercepts
( 255 − 5 105 52 , 0 ) , ( 255 + 5 105 52 , 0 ) (\dfrac{255-5\sqrt{105}}{52}, 0), (\dfrac{255+5\sqrt{105}}{52}, 0) ( 52 255 − 5 105 , 0 ) , ( 52 255 + 5 105 , 0 )
(d) the y-intercept
( 0 , 24 ) (0,24) ( 0 , 24 )
(e) intervals on which f is increasing
( 255 52 , ∞ ) (\dfrac{255}{52}, \infin) ( 52 255 , ∞ )
(f) intervals on which f is decreasing
( 0 , 255 52 ) (0, \dfrac{255}{52}) ( 0 , 52 255 )
(g) intervals on which f is constant
( − ∞ , 0 ) (-\infin, 0) ( − ∞ , 0 )
(h) the number at which f has a relative minimum
255 52 \dfrac{255}{52} 52 255
(i) the relative minimum of f
− 105 104 -\dfrac{105}{104} − 104 105
(j) f(0)
24 24 24
(k) The values of x for which f(x)=3
255 − 5 417 52 , 255 + 5 417 52 \dfrac{255-5\sqrt{417}}{52}, \dfrac{255+5\sqrt{417}}{52} 52 255 − 5 417 , 52 255 + 5 417
(l) Is f even, odd or neither?
The function f is neither even nor odd.
Comments