[size=150]Evaluate each expression for [math]a=9[/math], [math]b=-5[/math], and [math]c=-2[/math][br][/size][br][math]-b[/math]
[math]-b\pm\sqrt{a}[/math]
[list][size=150][*]Solve 1–2 equations by using the quadratic formula.[br][/*][*]Then, find and describe the error(s) in the worked solutions of the same equations as the ones you solved.[/*][/size][/list][br][table][tr][td]Equation 1: [math]\quad2x^2+3=8x[/math][/td][td]Equation 2: [math]\quad x^2+3x=10[/math][/td][/tr][tr][td]Equation 3: [math]\quad9x^2-2x-1=0[/math][/td][td]Equation 4: [math]\quad x^2-10x+23=0[/math][/td][/tr][/table][size=150][br]Here are the worked solutions with errors:[/size][br][br][table][tr][td]Equation 1:[math]\quad2x^2+3=8x[/math][/td][td]Equation 2:[math]\quad x^2+3x=10[/math][/td][/tr][tr][td][math]a=2,\, b= \text-8,\, c=3[/math][br][br][math]x=\frac{\text{-}b \pm \sqrt{b^2-4ac}}{2a} \\[br]x=\frac{\text{-}(\text{-}8) \pm \sqrt{(\text{-}8)^2-4(2)(3)}}{2(2)}\\ [br]x=\frac{8 \pm \sqrt{64-24}}{4}\\[br]x=\frac{8 \pm \sqrt{40}}{4}\\[br]x =2 \pm \sqrt{10}[/math][br][/td][td][math]a=1,b=3,c=10[/math][br][br][math]x=\frac{\text{-}b \pm \sqrt{b^2-4ac}}{2a}\\ [br]x=\frac{\text{-}3 \pm \sqrt{3^2-4(1)(10)}}{2(1)}\\ [br]x=\frac{\text{-}3 \pm \sqrt{9-40}}{2}\\ [br]x=\frac{\text{-}3 \pm \sqrt{\text{-}31}}{2}\\ [br]\text{No solutions}[/math][br][/td][/tr][tr][td][br]Equation 3:[math]\quad9x^2-2x-1=0[/math][/td][td][br]Equation 4:[math]\quad x^2-10x+23=0[/math][/td][/tr][tr][td][math]a = 9,\,b = \text-2,\, c = \text-1[/math][br][br][br][math]x=\frac{\text{-}b \pm \sqrt{b^2-4ac}}{2a}\\ [br]x=\frac{2 \pm \sqrt{(\text{-}2)^2-4(9)(\text{-}1)}}{2}\\ [br]x=\frac{2 \pm \sqrt{4+36}}{2}\\[br]x=\frac{2 \pm \sqrt{40}}{2}[/math][/td][td][math]a = 1,\,b = \text-10,\, c = 23[/math][br][br][math]x=\frac{\text{-}b \pm \sqrt{b^2-4ac}}{2a}\\ [br]x=\frac{\text{-}10 \pm \sqrt{(\text{-}10)^2-4(1)(23)}}{2}\\ [br]x=\frac{\text{-}10 \pm \sqrt{\text{-}100-92}}{2}\\[br]x=\frac{\text{-}10 \pm \sqrt{\text{-}192}}{2}\\ [br]\text{No solutions}[/math][/td][/tr][/table]
[size=150]The equation [math]h\left(t\right)=2+30t-5t^2[/math] represents the height, as a function of time, of a pumpkin that was [br]catapulted up in the air. Height is measured in meters and time is measured in seconds.[/size][br][br][size=100]The pumpkin reached a maximum height of 47 meters. How many seconds after launch did that happen? Show your reasoning.[/size]
Suppose someone was unconvinced by your solution. Find another way (besides the steps you already took) to show your solution is correct.[br]
[size=150]The equation [math]r\left(p\right)=80p-p^2[/math] models the revenue a band expects to collect as a function of the price of one concert ticket. Ticket prices and revenues are in dollars.[br][/size][br][size=100]A band member says that a ticket price of either $15.50 or $74.50 would generate approximately $1,000 in revenue. Do you agree? Show your reasoning.[/size]
[size=150]Function [math]g[/math] is defined by the equation [math]g\left(t\right)=2+30t-5t^2-47[/math]. Its graph opens downward.[br][/size][br]Find the zeros of function [math]g[/math] without graphing. Show your reasoning.
Explain or show how the zeros you found can tell us the vertex of the graph of [math]g[/math].[br]
Study the expressions that define functions [math]g[/math] and [math]h[/math] (which defined the height of the pumpkin). Explain how the maximum of function [math]h[/math], once we know it, can tell us the maximum of [math]g[/math].[br]