**Solutions of a Quadratic Equation on a Graph**

Hello, and welcome to this video about solutions of a **quadratic on a graph**! Today we’ll learn how to find solutions to a quadratic function by looking at its graph. We’ll also talk about how to graph a quadratic equation and analyze the graph to find solutions.

Before we get started, let’s review a few things. First, a **quadratic function** is a polynomial function, and its highest degree term is of the second degree. The graph of a quadratic function is a two-dimensional curve called a **parabola**. The parabola can open upward or downward and can vary in width. However, all parabolas share the same U-shape.

A parabola is symmetric over an invisible line called the **axis of symmetry**. The **vertex** is the point on the parabola where the graph intersects its axis of symmetry. In the parabola shown, \(y=x^{2}\), the vertex is the point at the bottom of the U-shaped parabola, seen here at point (0,0).

Notice that this parabola is symmetrical with respect to the y-axis, which is the axis of symmetry on this graph. The vertex is the point at the bottom of the U-shape.

If the graph of a quadratic function crosses the x-axis at two points, then the equation has two real rational solutions. These solutions are also called x-**intercepts** or **roots**. If it touches the x-axis at one point, it has one real rational solution. If the graph does not intersect the x-axis, then the equation has no real solutions.

A **quadratic equation** in standard form is written as \(ax^{2}+bx+c=0\), where \(a\neq 0\) and \(a\), \(b\), and \(c\) are all real numbers. We can solve a quadratic equation by **factoring**, **completing the square**, using the **quadratic formula**, or analyzing the graph of its function.

Consider the graph for \(y=x^{2}+x-6\). We can use this graph to solve the equation \(x^{2}+x-6=0\). Notice that \(y\) is replaced with \(0\). Replacing \(y\) with \(0\) indicates that we are solving for \(x\) by identifying the x-intercepts.

Looking at the graph, we can see that this quadratic function has two solutions. The graph intersects the \(x\)-axis at the points (-3,0) and (2,0), so the solutions for \(x\) are -3 and 2. In this example, there are two real rational solutions. The graph intersects the \(x\)-axis exactly two times.

Let’s look at another example. Consider the graph for \(y=-x^{2}+4\). Notice that the value for a is negative, so this parabola opens downward. We can use the graph to solve the equation \(-x^{2}+4=0\).

The graph intersects the \(x\)-axis at points (-2,0) and (2,0), so \(x=-2\), 2. In this example, there are two real rational solutions. The graph intersects the \(x\)-axis exactly two times.

Let’s try another one. Consider the graph for \(y=x^{2}-2x+1\). We can use the graph to solve the equation \(x^{2}-2x+1=0\).

Notice that this graph doesn’t pass through the \(x\)-axis. Instead, its vertex touches the \(x\)-axis at the point (1,0). In this case, there is only one real rational solution, \(x=1\).

Let’s try one more. Consider the graph for \(y=x^{2}+1\). We can use this graph to identify possible solutions to the equation \(x^{2}+1=0\).

Looking at this graph, we see that it doesn’t pass through the \(x\)-axis at all, which means there are no x-intercepts. In other words, there are no real solutions to this equation.

Quadratic functions occur in many real-world situations. Many questions about time, speed, and distance require quadratic equations to solve. An example of this is throwing a ball up into the air. Once a ball is thrown, it slows and eventually falls back down. When thrown, the ball’s path travels in an upside-down U-shape, or parabola. A quadratic equation can be used to find the position of the ball for the amount of time it’s in the air.

Another example is a boat traveling against a river current. If a boat travels 10 miles upstream and turns around to return to its starting point, its path will be in a U-shaped parabola. A quadratic equation can be used to find the boat’s speed and the length of the trip.

There are lots of other problem-solving instances in which quadratic equations are helpful. Understanding them helps us apply mathematical concepts in the real world. Let’s take a look at a story problem that we can solve by graphing a quadratic function.

The width of a rectangular field is 2 meters less than the length. The area is \(48\text{ m}^{2}\). Find the dimensions of the field.

First, start with the formula for the area of a rectangle, which is \(A=\text{ length}\times \text{ width}\). The word problem doesn’t give the field’s length, so use the variable \(x\) to represent length. Since the width is 2 meters less than the length, use the expression \(x-2\) to represent the width.

Next, substitute the values into the area formula. So we said:

\(\text{ A}=\text{ lw}\)

We were given that our area is equal to 48, and we said that length is \(x\) and width is \(x-2\).

\(48=x(x-2)\)

From here, simplify the equation and write it as a quadratic equation in standard form.

So the 48 is going to stay the same for right now and we’re going to distribute the \(x\) into the parentheses \((x-2)\). So \(x\cdot x=x^{2}\) and \(x\cdot (-2)=-2x\).

\(48=x^{2}-2x\)

Now remember, we want 0 on one side so we’re going to subtract 48 from both sides.

\(48-48=x^{2}-2x-48\)

That gives us:

\(0=x^{2}-2x-48\)

Now remember standard form technically has the 0 on the other side, so we’re just going to rewrite it as:

\(x^{2}-2x-48=0\)

Now we’re in proper standard form.

Now that we have the area expressed as a quadratic equation, we can create a graph to interpret possible solutions. To graph the parabola, we need the coordinates for its \(x\)-intercepts and its vertex.

Let’s start by identifying the coordinates for the \(x\)-intercepts. We can find these by factoring the equation \(x^{2}-2x-48=0\). Quadratic equations can be factored as \((x-p)(x-q)=0\), where \(p\) and \(q\) are solutions to the equation.

To factor the equation, identify two numbers with a sum of -2 (\(b\)) and a product of -48 (\(c\)). Two numbers with a sum of -2 and a product of -48 are -8 and 6. Therefore, we can factor this quadratic equation by writing it as \((x-8)(x+6)=0\). Now notice that earlier we said that it was \((x-p)(x-q)\), but here we have a plus sign. Well this is really the same as \((x-(-6))\), which then simplifies to \((x+6)\) because of the subtracting a negative.

Next, substitute values for \(x\) that make each equation true. Since \(8-8=0\), \(x=8\). Since \(-6+6=0\), \(x=-6\). The solutions for \(x\) are 8 and -6. So the coordinates for these \(x\)-intercepts are (8,0) and (-6,0). So the function passes through these two points.

Now that we’ve identified the coordinates of the \(x\)-intercepts, we can find the coordinates for the vertex. Start by finding the \(x\)-coordinate of the vertex. Since the vertex is the midpoint of the two x-intercepts, the vertex’s \(x\)-coordinate is the sum of the two \(x\)-intercepts divided by 2. So let’s do that. We have our two x-intercepts, 8 and -6, and to find the average of these two points, we’ll add them together and divide by 2.

\(\frac{8+(-6)}{2}=\frac{2}{2}=1\)

So the \(x\)-coordinate of our vertex is 1. To find the \(y\)-coordinate of the vertex, substitute 1 in for \(x\) into our original equation, which is \(y=x^{2}-2x-48\). So if we plug in 1 anywhere we see an \(x\), we’ll get:

\(y=(1)^{2}-2(1)-48\)

\(y=1-2-48\)

\(y=-1-48\)

\(y=-49\)

So our y-coordinate is equal to -49. Therefore, the vertex is at point (1,-49).

Now that we’ve identified the \(x\)-intercepts and the vertex, we can use these points to create a graph. Plot the points and join them by a smooth curve, as shown.

Now that we’ve graphed the function, let’s reconsider the context of the word problem. Recall that we are trying to identify the dimensions of a rectangular field. Since meters can’t be measured in negative numbers, the only solution for x that makes sense here is 8. Recall that x represents the field’s length, which is 8 meters. Recall that the width is 2 meters less than the length. Since \(8-2=6\), the width of the field is 6 meters. The field is 8 meters long and 6 meters wide.

Now it’s your turn. I’m going to give you a quadratic function and you need to factor it, identify the coordinates for its x-intercepts, identify the coordinates for its vertex, and graph it. This problem is challenging because it involves several steps, but I know you can handle it! The quadratic function is \(y=x^{2}-6x-16\). Pause the video here and try it yourself.

Let’s take a look at this quadratic function together. The first step is factoring the equation \(x^{2}-6x-16=0\). To do so, we need to identify two numbers that have a sum of -6 (\(b\)) and a product of -16 (\(c\)). Then, we’ll write these numbers into the factored equation \((x\text{ })(x\text{ })=0\). The correct numbers are -8 and 2 because \(-8\times 2=-16\), and \(-8+2=-6\). So we’re going to plug in -8 and 2:

\((x-8)(x+2)=0\)

Next, substitute a value for \(x\) in both subtraction problems that makes the equation true. Since \(8-8=0\), \(x=8\). Since \(-2+2=0\), \(x=-2\). Therefore, the solutions for \(x\) are 8 and -2. We can write the \(x\)-intercept values as the coordinate points (8,0) and (-2,0).

Now that we’ve identified the \(x\)-intercepts, we need to find the coordinates for the vertex, starting with its \(x\)-coordinate. Recall that the \(x\)-coordinate of the vertex is the midpoint of the \(x\)-intercepts. So for our vertex, to find the \(x\)-coordinate all we have to do is add these two values and divide by 2.

\(\frac{8+(-2)}{2}=\frac{6}{2}=3\)

So the \(x\)-coordinate of the vertex is 3. To find the \(y\)-coordinate, substitute 3 for \(x\) into our original equation \(y=x^{2}-6x-16\):

\(y=(3)^{2}-6(3)-16\)

\(y=9-18-16\)

\(y=-9-16\)

\(y=-25\)

Therefore, the vertex is at point (3,-25).

Now that we’ve identified the \(x\)-intercepts and the vertex, we can use points (8,0), (-2,0), and (3,-25) to create a graph. Plot the points and join them by a smooth curve.

Great job!

I hope this video about solutions of a quadratic on a graph was helpful. Thanks for watching, and happy studying!

## Practice Questions

**Question #1:**

Solve \(x^2+3x-28=0\) by using the graph shown below:

No real solution

**Answer:**

Looking at the graph, we see that the quadratic function crosses the \(x\)-axis at two points. This means that there are two real solutions to this equation. The graph intersects the \(x\)-axis at points \((-7,0)\) and \((4,0)\), so the solutions for \(x\) are –7 and 4.

**Question #2:**

Solve \(x^2+10x+25=0\) by using the graph shown below:

No real solution

**Answer:**

Looking at the graph, we see that the quadratic function touches the \(x\)-axis at exactly one point. This means that there is one real solution to this equation. The vertex touches the \(x\)-axis at point \((-5,0)\), so the solution for \(x\) is –5.

**Question #3:**

Solve \(x^2+x+4=0\) by using the graph shown below:

No real solutions

**Answer:**

Looking at the graph, we see that the quadratic function does not pass through the \(x\)-axis at all. This means that there are no real solutions to this equation.

**Question #4:**

The length of a rectangular flag is 4 feet longer than its width. The area is 96 square feet. Find the dimensions of the flag by writing a quadratic equation and graphing it on the coordinate plane.

The width is -12 feet, and the length is -8 feet.

The width is 8 feet, and the length is 12 feet.

The width is -2 feet, and the length is -100 feet.

The width is 2 feet, and the length is 100 feet.

**Answer:**

Start by recalling the formula for the area of a rectangle, which is length times width. Substitute the values from the word problem into this formula. Since the width is not known, use w to represent width. Since the length of the flag is 4 feet longer than its width, use \(w+4\) to represent the length.

\(A=lw\)

\(96=w\left(w+4\right)\)

Next, simplify the equation by distributing \(w\). Write the quadratic equation in standard form.

\(96=w^2+4w\)

\(96-96=w^2+4w-96\)

\(0=w^2+4w-96\)

\(w^2+4w-96=0\)

Then, identify the coordinates for the \(x\)-intercepts. Factor the equation by finding two numbers that result in a sum of 4 and a product of -96. These numbers are 12 and -8.

\(\left(w+12\right)\left(w-8\right)=0\)

From here, equate each binomial to 0 and solve for \(w\). The solutions are -12 or 8. The coordinates for these \(x\)-intercepts are \((-12,0)\) and \((8,0)\). The graph of the function passes through the \(x\)-axis at -12 and 8.

\(w+12=0\hspace{20 mm}w-8=0\)

\(w=-12\hspace{20 mm}w=8\)

Now that we know the \(x\)-intercepts, find the coordinates for the vertex, \((h,k)\). Start by finding \(h\), which is the vertex’s \(x\)-coordinate. Divide the sum of the \(x\)-intercepts by 2:

\(h=\frac{-12+8}{2}=-\frac{4}{2}=-2\)

Next, find \(k\), which is the vertex’s \(y\)-coordinate. Substitute -2 into the quadratic equation for \(w\) and simplify:

\(k=(-2)^2+4\left(-2\right)-96\)

\(k=4-8-96\)

\(k=-100\)

The vertex of this function is \((-2,-100)\). Plot the vertex and the \(x\)-intercepts onto the coordinate plane and join the points with a smooth curve.

Since this problem is about the area of a flag, the negative solution, -12, does not apply to this scenario. Therefore, the width of the flag is 8 feet. Since the length of the flag is 4 feet longer than its width, the length of the flag is 12 feet.

**Question #5:**

Nicole drops a ball from 4 feet above the ground and watches as it bounces back into the air. The path of Nicole’s ball takes the shape of a parabola and can be represented by the equation \(\left(x-2\right)^2=0\). Graph this quadratic function to figure out the location of where the ball hits the ground in relation to where Nicole is standing.

Nicole’s ball hits the ground 2 feet from where she is standing.

Nicole’s ball hit the ground 4 feet from where she is standing.

Nicole’s ball hit the ground -2 feet from where she is standing.

Nicole’s ball hit the ground -4 feet from where she is standing.

**Answer:**

Start by writing the quadratic equation as two binomials.

\(\left(x-2\right)^2=0\)

\(\left(x-2\right)\left(x-2\right)=0\)

Next, set each binomial equal to 0 and solve for \(x\). Since both binomials are the same, there is one solution, which is 2. The graph touches the \(x\)-axis at \((2,0)\).

\(x-2=0\)

\(x=2\)

Then, identify the \(y\)-intercept. Since the value of \(c\) in the expanded equation is 4, the \(y\)-intercept is 4. Nicole drops the ball from 4 feet above the ground, so the point representing the \(y\)-intercept is \((0,4)\).

\(\left(x-2\right)\left(x-2\right)=0\)

\(x^2-2x-2x+4=0\)

\(x^2-4x+4=0\)

\(c=4\)

From here, plot the \(x-\) and \(y\)-intercepts onto the coordinate plane and join the points with a smooth curve:

Based on this information, Nicole stands at point \((0,0)\) and drops the ball from 4 feet above ground at point \((0,4)\). The ball hits the ground at point \((2,0)\). Therefore, the ball hits the ground 2 feet away from where Nicole stands.