McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
3. The Quadratic Formula and the Discriminant
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Exercise 33 Page 195

Practice makes perfect
a We are given an equation that models the position of the video game cartridge. The graph below illustrates the movement. There is also a horizontal line on the graph, which represents the height of the window.

We can see from the graph that the toss is not strong enough to reach the level of the window. To confirm this, let's find the vertex of the parabolic path. We know that this vertex is on the axis of symmetry. 2 &Quadratic function: &&f(x)= ax^2+ bx+ c &Axis of symmetry:&&x=-b/2 a &Vertex:&&(-b/2 a,f(-b/2 a)) Let's identify the coefficients in our quadratic expression. - 16t^2+ 35t+ 5 We see that a= - 16, b= 35, and c= 5. Let's substitute these values into the formula above.

-b/2a
-35/2( - 16)
-35/- 32
35/32

This means that the cartridge reaches the maximum position after 3532 seconds. To find the maximum height, we can substitute this value in the expression for the height.

h(t)=- 16t^2+35t+5
h( 35/32)=- 16( 35/32)^2+35( 35/32)+5
â–¼
Simplify right-hand side
h(35/32)=- 16* 1225/1024+35*35/32+5
h(35/32)=- 1225/64+1225/32+5
h(35/32)=5+1225/64
h(35/32)=5+19 964
h(35/32)=24 964
h(35/32)=24.140625
h(35/32)≈ 24.14

This confirms what we see on the graph. The highest point of the path of the cartridge is below the window. Darnell will have 0 chance to catch it.

b To find the time when the cartridge hits the ground, let's look at the graph again. This time we focus on the horizontal axis intercept, which corresponds to 0 height.The first coordinate of this point gives the time we are looking for.

To find this axis intercept, we need to find the solution of h(t)=0, which is a quadratic equation. - 16t^2+35t+5=0We will use the Quadratic Formula. 2 &Quadratic equation: && ax^2+ bx+ c=0 &Solutions:&&x=- b±sqrt(b^2-4 a c)/2 a Let's identify the coefficients in our quadratic expression. - 16t^2+ 35t+ 5 We see that a= - 16, b= 35, and c= 5. Let's substitute these values into the Quadratic Formula.

t=- b±sqrt(b^2-4ac)/2a
t=- 35±sqrt(35^2-4( -16)( 5))/2( -16)
â–¼
Simplify right-hand side
t=- 35±sqrt(1225-4( -16)( 5))/2( -16)
t=- 35±sqrt(1225-20( -16))/2( -16)
t=- 35±sqrt(1225-(-320))/-32
t=- 35±sqrt(1225+320)/-32
t=-35±sqrt(1545)/-32

Now we use a calculator to find approximate values of these solutions. -35+sqrt(1545)/-32&≈ -0.13 -35-sqrt(1545)/-32&≈ 2.32 Since the solution we are looking for represents time, we need the positive solution. The cartridge will hit the ground approximately 2.32 seconds after Jack tossed it.