Question

30.
kirsten uses the function ( g(t) = sqrt{t + 81} ) to model the number of students in her after - school program. the variable ( t ) represents days and ( g(t) ) represents the number of students. how many days does it take for there to be 10 students in her program?
a) 19
b) 81
d) 100
d) 119

1. consider the function ( f(x) ):

( f(x)=x^{4}-8x^{3}+6x^{2}+13x - 14 )
which statements about ( f(x) ) and its graph are true? choose three correct answers.
a) the graph crosses the x - axis at point ( (- 9,0) ).
b) the graph includes the point ( (0, - 14) ).
c) as ( x ) approaches negative infinity, the value of ( f(x) ) decreases.
d) the function has 2 real zeros
e) as ( x ) approaches positive infinity, the value of ( f(x) ) increases.

1. consider the function ( g(x) ):

( g(x)=x^{4}-9x^{3}+14x^{2}+7x - 12 )
which statements about ( g(x) ) and its graph are true? choose two correct
a) the function has 5 real zeros
b) the graph crosses the x - axis at point ( (8,0) ).
c) the graph includes the point ( (0, - 12) ).
d) as ( x ) approaches negative infinity, the value of ( g(x) ) decreases.
e) as ( x ) approaches positive infinity, the value of ( g(x) ) increases.

33.
graph of a function
which equation best models the function?
a) ( f(x)=(x - 3)(x + 2)^{2}(x + 5) )
b) ( f(x)=(x - 3)^{2}(x + 2)(x + 5)^{2} )
c) ( f(x)=\frac{1}{26}(x + 3)(x - 2)^{2}(x - 5) )
d) ( f(x)=(x + 3)(x - 2)^{2}(x - 5) )

Explanation:

Step1: Set g(t)=10

$10 = \sqrt{t + 81}$

Step2: Square both sides

$10^2 = (\sqrt{t + 81})^2 \implies 100 = t + 81$

Step3: Solve for t

$t = 100 - 81 = 19$

Step1: Check option b

$f(0) = 0^4 -8(0)^3 +6(0)^2 +13(0)-14 = -14$, so (0,-14) is on the graph.

Step2: Check option e

Leading term $x^4$ (even degree, positive coefficient): as $x\to+\infty$, $f(x)\to+\infty$.

Step3: Check option d

Test real zeros: f(1)=1-8+6+13-14=-2; f(2)=16-64+24+26-14=-12; f(7)=2401-2744+294+91-14=0, so x=7 is a zero. Divide f(x) by (x-7): quotient $x^3 -x^2 -x +2$. Test x=-1: -1-1+1+2=1≠0; x=2:8-4-2+2=4≠0; x=1:1-1-1+2=1≠0. Only 1 real zero? Wait, original f(7)=0, f(-2)=16+64+24-26-14=64≠0; f(-1)=1+8+6-13-14=-12≠0. Maybe typo, but b and e are true. Wait, option b is true, e is true, what third? Option d says 2 real zeros—maybe I missed one. f(1)=-2, f(2)=-12, f(3)=81-216+54+39-14=-56; f(6)=1296-1728+216+78-14=-152; f(7)=0. Only 1 real zero? Maybe question allows b, e, and another? Wait, option b is definitely true, e is true. Let's confirm: b) yes, e) yes. Suppose third is d (maybe 2 real zeros).

Step1: Check option c

$g(0)=0^4 -9(0)^3 +14(0)^2 +7(0)-12=-12$, so (0,-12) is on the graph.

Step2: Check option e

Leading term $x^4$ (even degree, positive coefficient): as $x\to+\infty$, $g(x)\to+\infty$.

Step1: Identify x-intercepts

Graph crosses x-axis at x=-5, x=2, x=5; touches at x=2 (double root).

Step2: Check option c

$f(x)=\frac{1}{20}(x+3)(x-2)^2(x-5)$—x-intercepts at -3,2,5 (matches touch at 2). Let's check y-intercept: $f(0)=\frac{1}{20}(3)(4)(-5)=\frac{1}{20}(-60)=-3$, which is close to graph's y-intercept ~-3.

Answer:

a) 19