A backyard farmer wants to enclose a rectangular space for a new garden. She has purchased [latex]80[/latex] feet of wire fencing to enclose three sides and will put the fourth side against the backyard fence. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[/latex].

In a scenario like this involving geometry, it is often helpful to draw a picture. It might also be helpful to introduce a temporary variable, [latex]W[/latex], to represent the side of fencing parallel to the fourth side or backyard fence.

Since we know we only have [latex]80[/latex] feet of fence available, we know that [latex]L+W+L=80[/latex], or more simply, [latex]2L+W=80[/latex]. This allows us to represent the width, [latex]W[/latex], in terms of [latex]L[/latex]: [latex]W=80-2L[/latex].

Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so [latex]A=LW=L(80-2l)[/latex], so [latex]A(L)=80L-2L^2.[/latex] This formula represents the area of the fence in terms of the variable length [latex]L[/latex].

Write an equation for the quadratic graphed below as a transformation of [latex]f(x)=x^2[/latex].

We can see the graph is the basic quadratic shifted to the left [latex]2[/latex] and down [latex]3[/latex], putting the vertex at [latex](-2, -3)[/latex], giving a formula in the form [latex]g(x)=a(x+2)^2-3[/latex]. By plugging in a point that falls on the grid, such as [latex](0, -1)[/latex], we can solve for the stretch factor:
[latex]\begin{align*} -1=&a(0+2)^2-3 \\ 2=& 4a \\ a=& \frac{1}{2} \end{align*}[/latex]

The equation for this formula is [latex]g(x)=\frac{1}{2}(x+2)^2-3.[/latex]

Find the vertical and horizontal intercepts of the quadratic [latex]f(x)=3x^2+5x-2[/latex].

We can find the vertical intercept by evaluating the function at an input of zero: [latex]f(0)=3(0)^2+5(0)-2=-2.[/latex] So the vertical intercept is at [latex](0, -2)[/latex].

For the horizontal intercepts, we solve for when the output will be zero: [latex]0=3x^2+5x-2.[/latex] In this case, the quadratic can be factored easily, providing the simplest method for solution: [latex]0=(3x-1)(x+2),[/latex] so either
[latex]\begin{align*} 0=& 3x-1\\ x=& \frac{1}{3} \end{align*}[/latex]
or
[latex]\begin{align*} 0=& x+2\\ x=& -2 \end{align*}[/latex]
So the horizontal intercepts are at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex](-2,0)[/latex].

A ball is thrown upward from the top of a [latex]40[/latex]-foot-high building at a speed of [latex]80[/latex] feet per second. The ball’s height above ground can be modeled by the equation [latex]H(t)=-16t^2+80t+40.[/latex] When does the ball hit the ground?

To find when the ball hits the ground, we need to determine when the height is zero, i.e., when [latex]H(t) = 0[/latex]. While we could do this using the transformation form of the quadratic, we can also use the quadratic formula: [latex]t=\frac{-80\pm \sqrt{80^2-4(-16)(40)}}{2(-16)}=\frac{-80\pm\sqrt{8960}}{-32}.[/latex]

Since the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions: [latex]t=\frac{-80-\sqrt{8960}}{-32}\approx 5.458 \quad\text{or}\quad t=\frac{-80+\sqrt{8960}}{-32}\approx -0.458.[/latex]

The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about [latex]5.458[/latex] seconds.