Question

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math ii unit 4 exam #4
unit 4: solving quadratic equations
q1.) a football is kicked from the ground and travels at a rate of 18 meters per second. the function represents the ball’s height above the ground at t seconds. based on this function model: ( f(x) = 18x - 6x^2 )
what is football’s highest point and how long did it take to get there?
q2.) solve by factoring: (mcas q)
( 5x - 1 = x^2 + 2x + 1 )
q3) using the factoring method solve for this quadratic equation. ( x^2 - 10x + 24 = 0 )
q.4) given the quadratic equation in factored form solve for x? (mcas q) ( 2x(3x - 4) = 0 )
q.5) solve for by completing the square: ( x^2 + 8x + 16 = 100 )
q.6) use any method to find the solution to the equation ( x^2 + 14x + 49 = 36 ).
q.7) solve the equation using the factoring method: (mcas q)
( 6x^2 + 18x = 0 )
q.8) given the graph find the simplest function rule. assume ( a = 1 ).
graph of a parabola with vertex at (0,7) and roots at (-3,0) and (3,0)
q.9) find the exact function rule of this graph using the formula ( y = a(x + m)(x + n) ) where ( m, n ) are the roots.
graph of a parabola with vertex at (0,7) and roots at (-3,0) and (3,0)
q.10) solve for x:
( (x + 5)^2 = 4 )

Explanation:

Q1:

Step1: Find vertex x-coordinate

For $f(x)=ax^2+bx+c$, vertex $x=-\frac{b}{2a}$. Here $a=-6, b=18$:
$x=-\frac{18}{2(-6)} = \frac{18}{12} = 1.5$

Step2: Find max height at x=1.5

Substitute $x=1.5$ into $f(x)$:
$f(1.5)=18(1.5)-6(1.5)^2 = 27 - 6(2.25) = 27 - 13.5 = 13.5$

Q2:

Step1: Rearrange to standard form

Bring all terms to one side:
$x^2+2x+1-5x+1=0 \implies x^2-3x+2=0$

Step2: Factor the quadratic

Find two numbers that multiply to 2 and add to -3:
$(x-1)(x-2)=0$

Step3: Solve for x

Set each factor to 0:
$x-1=0 \implies x=1$; $x-2=0 \implies x=2$

Q3:

Step1: Factor the quadratic

Find two numbers that multiply to 24 and add to -10:
$(x-4)(x-6)=0$

Step2: Solve for x

Set each factor to 0:
$x-4=0 \implies x=4$; $x-6=0 \implies x=6$

Q4:

Step1: Set each factor to 0

Use zero product property:
$2x=0$ or $3x-4=0$

Step2: Solve each equation

$2x=0 \implies x=0$; $3x=4 \implies x=\frac{4}{3}$

Q5:

Step1: Rewrite as perfect square

Left side is a perfect square:
$(x+4)^2=100$

Step2: Take square roots

$\sqrt{(x+4)^2}=\pm\sqrt{100} \implies x+4=\pm10$

Step3: Solve for x

$x=-4+10=6$; $x=-4-10=-14$

Q6:

Step1: Rewrite as perfect square

Left side is a perfect square:
$(x+7)^2=36$

Step2: Take square roots

$\sqrt{(x+7)^2}=\pm\sqrt{36} \implies x+7=\pm6$

Step3: Solve for x

$x=-7+6=-1$; $x=-7-6=-13$

Q7:

Step1: Factor out GCF

Factor $6x$ from the left side:
$6x(x+3)=0$

Step2: Solve for x

Set each factor to 0:
$6x=0 \implies x=0$; $x+3=0 \implies x=-3$

Q8:

Step1: Identify roots and vertex

Roots are $x=-3$ and $x=3$; vertex is $(0,7)$

Step2: Write factored form

With $a=-1$, use roots:
$y=-(x+3)(x-3)$

Step3: Simplify to standard form

Expand the expression:
$y=-(x^2-9) = -x^2+7$

Q9:

Step1: Identify roots

Roots are $x=-3$ and $x=3$

Step2: Substitute into factored form

With $a=-1$:
$y=-1(x+3)(x-3)$

Q10:

Step1: Take square roots

$\sqrt{(x+5)^2}=\pm\sqrt{4} \implies x+5=\pm2$

Step2: Solve for x

$x=-5+2=-3$; $x=-5-2=-7$

Answer:

Q1: Highest point is 13.5 meters, reached at 1.5 seconds
Q2: $x=1$, $x=2$
Q3: $x=4$, $x=6$
Q4: $x=0$, $x=\frac{4}{3}$
Q5: $x=6$, $x=-14$
Q6: $x=-1$, $x=-13$
Q7: $x=0$, $x=-3$
Q8: $y=-x^2+7$
Q9: $y=-(x+3)(x-3)$
Q10: $x=-3$, $x=-7$