Example: Finding the y- and x-Intercepts of a Parabola

Find the [latex]y[/latex]- and [latex]x[/latex]-intercepts of the quadratic [latex]f\left(x\right)=3{x}^{2}+5x - 2[/latex].

Answer: We find the [latex]y[/latex]-intercept by evaluating [latex]f\left(0\right)[/latex].

[latex]f\left(0\right)=3{\left(0\right)}^{2}+5\left(0\right)-2=-2[/latex]

So the [latex]y[/latex]-intercept is at [latex]\left(0,-2\right)[/latex]. For the [latex]x[/latex]-intercepts, or roots, we find all solutions of [latex]f\left(x\right)=0[/latex].

[latex]3{x}^{2}+5x - 2=0[/latex]

In this case, the quadratic can be factored easily, providing the simplest method for solution.

[latex]\left(3x - 1\right)\left(x+2\right)=0[/latex] [latex-display]\begin{align}&3x - 1=0&x+2=0\end{align}[/latex-display] [latex]x=\frac{1}{3}\hspace{8mm}[/latex] or [latex]\hspace{8mm}x=-2[/latex]

So the roots are at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].

Analysis of the Solution

By graphing the function, we can confirm that the graph crosses the [latex]y[/latex]-axis at [latex]\left(0,-2\right)[/latex]. We can also confirm that the graph crosses the [latex]x[/latex]-axis at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].

Example: Finding the Roots of a Parabola

Find the [latex]x[/latex]-intercepts of the quadratic function [latex]f\left(x\right)=2{x}^{2}+4x - 4[/latex].

Answer: We begin by solving for when the output will be zero.

[latex]0=2{x}^{2}+4x - 4[/latex]

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

[latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]

We know that [latex]a=2[/latex]. Then we solve for [latex]h[/latex] and [latex]k[/latex].

[latex]\begin{align}&h=-\dfrac{b}{2a}=-\dfrac{4}{2\left(2\right)}=-1\\[1mm]&\end{align}[/latex]

[latex]\begin{align}k&=f\left(h\right)\\&=f\left(-1\right)\\&=2{\left(-1\right)}^{2}+4\left(-1\right)-4\\&=-6\end{align}[/latex]

So now we can rewrite in standard form.

[latex]f\left(x\right)=2{\left(x+1\right)}^{2}-6[/latex]

We can now solve for when the output will be zero.

[latex]\begin{align}&0=2{\left(x+1\right)}^{2}-6 \\ &6=2{\left(x+1\right)}^{2} \\ &3={\left(x+1\right)}^{2} \\ &x+1=\pm \sqrt{3} \\ &x=-1\pm \sqrt{3} \end{align}[/latex]

The graph has [latex]x[/latex]-intercepts at [latex]\left(-1-\sqrt{3},0\right)[/latex] and [latex]\left(-1+\sqrt{3},0\right)[/latex].

Analysis of the Solution

We can check our work by graphing the given function on a graphing utility and observing the roots.

Example: Solving a Quadratic Equation with the Quadratic Formula

Solve [latex]{x}^{2}+x+2=0[/latex].

Answer: Let’s begin by writing the quadratic formula: [latex]x=\dfrac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]. When applying the quadratic formula, we identify the coefficients [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex]. For the equation [latex]{x}^{2}+x+2=0[/latex], we have [latex]a=1[/latex], [latex]b=1[/latex], and [latex]c=2[/latex]. Substituting these values into the formula we have:

[latex]\begin{align}x&=\dfrac{-b\pm \sqrt{{b}^{2}-4ac}}{2a} \\[1.5mm] &=\dfrac{-1\pm \sqrt{{1}^{2}-4\cdot 1\cdot \left(2\right)}}{2\cdot 1} \\[1.5mm] &=\dfrac{-1\pm \sqrt{1 - 8}}{2} \\[1.5mm] &=\dfrac{-1\pm \sqrt{-7}}{2} \\[1.5mm] &=\dfrac{-1\pm i\sqrt{7}}{2} \end{align}[/latex]

The solutions to the equation are [latex]x=\dfrac{-1+i\sqrt{7}}{2}[/latex] and [latex]x=\dfrac{-1-i\sqrt{7}}{2}[/latex] or [latex]x=-\dfrac{1}{2}+\dfrac{i\sqrt{7}}{2}[/latex] and [latex]x=-\dfrac{1}{2}-\dfrac{i\sqrt{7}}{2}[/latex].

Analysis of the Solution

This quadratic equation has only non-real solutions.

Example: Applying the Vertex and x-Intercepts of a Parabola

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

Answer: a. The ball reaches the maximum height at the vertex of the parabola.

[latex]\begin{align}h&=-\dfrac{80}{2\left(-16\right)} \\[1mm]&=\dfrac{80}{32} \\[1mm] &=\dfrac{5}{2} \\[1mm] &=2.5 \end{align}[/latex]

The ball reaches a maximum height after 2.5 seconds. b. To find the maximum height, find the [latex]y[/latex] coordinate of the vertex of the parabola.

[latex]\begin{align}k&=H\left(2.5\right) \\[1mm] &=-16{\left(2.5\right)}^{2}+80\left(2.5\right)+40 \\[1mm] &=140\hfill \end{align}[/latex]

The ball reaches a maximum height of 140 feet. c. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\left(t\right)=0[/latex]. We use the quadratic formula.

[latex]\begin{align} t&=\dfrac{-80\pm \sqrt{{80}^{2}-4\left(-16\right)\left(40\right)}}{2\left(-16\right)} \\[1mm] &=\dfrac{-80\pm \sqrt{8960}}{-32} \end{align}[/latex]

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

[latex]t=\dfrac{-80+\sqrt{8960}}{-32}\approx 5.458\hspace{3mm}[/latex] or [latex]\hspace{3mm}t=\dfrac{-80-\sqrt{8960}}{-32}\approx -0.458[/latex]

Since the domain starts at [latex]t=0[/latex] when the ball is thrown, the second answer is outside the reasonable domain of our model. The ball will hit the ground after about 5.46 seconds.