Question

exit ticketsolving quadratic equations using graphs and tables1. use the graph to find the solutions of$-\frac{1}{2}x^{2}-x + 4 = 0$.2. solve the equation $x^{2}-2x - 3 = 0$ by graphing3. the path of a volleyball thrown over a net is modeled with the function$a(x) = -0.02x^{2}+0.6x + 5$, where x is the horizontal distance, in feet, from thestarting point, and a is the altitude of the ball in feet. write an equation thatcan be solved to find out how far the ball travels horizontally before it hits theground. then find the distance rounded to the nearest foot.equation: _________________________ distance: __________1. what is the factored form of the function shownin the graph?a $(x + 4)(x - 1)$b $(x - 4)(x + 1)$c $(x - 4)(x - 1)$d $(x + 4)(x + 1)$2. what are the solutions of the equation$3x^{2}-5x + 2 = 0$?a 1 and 2b 1 and $\frac{2}{3}$c -1 and -2d -1 and $\frac{2}{3}$

Explanation:

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Problem 1 (Top Section, #1):

Step1: Identify x-intercepts (solutions)

The solutions of $-\frac{1}{2}x^2 - x + 4 = 0$ are the x-values where the graph crosses the x-axis. By analyzing the quadratic:

1. Rewrite in standard form: $-\frac{1}{2}x^2 - x + 4 = 0$
2. Multiply by -2: $x^2 + 2x - 8 = 0$
3. Factor: $(x+4)(x-2)=0$

Solutions: $x=-4, x=2$

Problem 2 (Top Section, #2):

Step1: Find x-intercepts of $x^2-2x-3=0$

Factor the quadratic: $(x-3)(x+1)=0$
Solutions: $x=3, x=-1$

Problem 3 (Top Section, #3):

Step1: Set altitude to 0 (ground level)

When the ball hits the ground, $A(x)=0$, so the equation is:
$$-0.02x^2 + 0.6x + 5 = 0$$

Step2: Solve using quadratic formula

For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here, $a=-0.02$, $b=0.6$, $c=5$
$$x=\frac{-0.6\pm\sqrt{0.6^2-4(-0.02)(5)}}{2(-0.02)}$$
$$x=\frac{-0.6\pm\sqrt{0.36+0.4}}{-0.04}=\frac{-0.6\pm\sqrt{0.76}}{-0.04}$$
$$x=\frac{-0.6\pm0.8718}{-0.04}$$
Take the positive solution (distance cannot be negative):
$$x=\frac{-0.6-0.8718}{-0.04}=\frac{-1.4718}{-0.04}\approx37$$

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Problem 1 (Bottom Multiple Choice):

Step1: Identify x-intercepts from graph

The graph crosses the x-axis at $x=-1$ and $x=4$.

Step2: Match to factored form

If $x=r$ is a root, the factor is $(x-r)$. So factors are $(x-4)$ and $(x+1)$.

Problem 2 (Bottom Multiple Choice):

Step1: Factor the quadratic $3x^2-5x+2=0$

$$3x^2-3x-2x+2=0$$
$$3x(x-1)-2(x-1)=0$$
$$(3x-2)(x-1)=0$$

Step2: Solve for x

Set each factor to 0:
$3x-2=0 \implies x=\frac{2}{3}$
$x-1=0 \implies x=1$

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Answer:

1. $x=-4$ and $x=2$
2. $x=-1$ and $x=3$
3. Equation: $\boldsymbol{-0.02x^2 + 0.6x + 5 = 0}$; Distance: $\boldsymbol{37}$ feet

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1. B. $(x - 4)(x + 1)$
2. B. 1 and $\frac{2}{3}$