Quadratic Function

Quadratic functions are an essential part of any algebra course. A thorough understanding of quadratic functions is necessary if we want to pursue advanced mathematical subjects like calculus. Quadratic functions can be used to model many different real-world phenomena. For instance, if we have a rocket that we want to launch, its expected trajectory could be modeled with a quadratic equation. Quadratic equations can also be used in economics and finance.

How to work with quadratic functions

A quadratic equation takes the general form of

$f\left(x\right)={ax}^2+bx+c$

When we graph a quadratic function, it takes the form of a parabola, which is a two-dimensional curve. Below is an example of a parabola in the form of $y=x^2$ . This is a basic quadratic function where the b and c values both equal zero.

We can alter the front coefficient a to make the graph thinner or broader. We can even flip the graph upside down by making the a coefficient negative. Check it out.

Notice that when the coefficient a gets smaller, the parabola gets wider. Likewise, when it gets larger, the graph gets thinner. The vertex of the parabola is the U-shaped area where it changes direction. When writing equations for parabolas, we can also write them in vertex form.

$y=a{(x–h)}^2+k$

When using this equation, remember that the parabola's vertex is at (h, k).

In order to find the x-coordinate of the vertex from the standard form equation, we use the expression

$\frac{-b}{2a}$

Let's attempt a practice problem.

If we have a quadratic equation defined by $y=x^2-6x+11$ , where does the vertex lie?

In this quadratic equation, $a=1$ and $b=-6$ . Therefore, we can plug these into our formula to find the x-coordinate of the vertex.

$\frac{-b}{2a}$

$\frac{-(-6)}{2\left(1\right)}$

$\frac{6}{2}=3$

Our x-coordinate of the vertex is 3. To get the y-coordinate, we simply plug this value into our quadratic function like so:

$y=3^2-6\left(3\right)+11$

$=9–18+11$

$=2$

The vertex of this quadratic function occurs at the point (3, 2). We can graph the function on paper to confirm this result.

Let's do another.

If we have the quadratic equation $y=2x^2+3x+1$ , where does the vertex lie?

From the standard form of this equation, we know that $a=2$ and $b=3$ . Now we can plug these into our formula to find the x-coordinate.

$\frac{-b}{2a}$

$\frac{-3}{\left(2\right)2}$

$\frac{-3}{4}$

This is our x-coordinate. Now, to get the y-coordinate, we need to plug this x-value into our equation.

$y=2{\left(\frac{-3}{4}\right)}^2+3\left(\frac{-3}{4}\right)+1$

$y=\frac{9}{8}+\left(-\frac{9}{4}\right)+1$

$y=-\frac{1}{8}$

Therefore, the vertex of our parabola will lie at the coordinate $(-\frac{3}{4},-\frac{1}{8})$ .

In math, we need to get used to working with messier numbers like in the problem above. Not every problem we encounter will yield clean whole numbers.

We may also be asked to calculate the axis of symmetry of a parabola. This is a vertical line that passes through the vertex of the parabola and splits it down the middle. We can calculate the axis of symmetry with the formula

$x=-\frac{b}{2a}$

This is the same formula for calculating the x-value of a vertex. This is because the axis of symmetry passes through the vertex by definition.

Find the axis of symmetry of a parabola defined by the equation $y=2x^2+x-1$ .

In this equation, $a=2$ and $b=1$ . Let's plug these into the formula.

$x=-\frac{b}{2a}$

$x=-\frac{1}{2}\times 2$

$x=-\frac{1}{4}$

Therefore, the axis of symmetry is the line $x=-\frac{1}{4}$ .

We may also be asked to calculate the intercepts of a quadratic equation. The y-intercept is a point at which the graph crosses the y-axis. We can calculate the y-intercept of a graph by plugging zeros into the x-values of the function. There are a few techniques for finding the x-intercept, including the quadratic formula, completing the square, and factoring. Bear in mind that x-intercepts may not exist depending on the shape of the graph.

Find the y-intercept of the function defined by $y=x^2-2x-3$ .

We'll find the y-intercept by inserting zero into the x-values.

$y=0^2-0-3$

$y=-3$

The function has a y-intercept of $(0,-3)$ .

In addition, we may have questions that ask us about the domain and range of a quadratic function. Recall that the domain is the set of all x-values that a function can accept. The range is the set of y-values that a function outputs. If we have a function like $y=x^2$ , it spans the full set of x-values. Therefore, the domain is $-\infty <x<\infty$ . These symbols indicate that the domain stretches from negative infinity to infinity. In interval notation, this is $(-\infty ,\infty )$ . The range of values starts at and includes $y=0$ , then continues up toward positive infinity. This can be expressed as $0\le y<\infty$ , or $\left[0,\mathrm{\infty }\right)$ . The closed bracket indicates that the function touches zero. The parenthesis indicates that there isn't an endpoint.

Quadratic function practice problems

1. Find the vertex of $y=x^2+4x-3$ .

We'll start by finding the x-value of the vertex with the formula $–\frac{b}{2a}$ .

$-\frac{4}{2\left(1\right)}$

$=-2$ .

Plug our x-value of 2 into the equation to get the y-coordinate.

$y={\left(-2\right)}^2+4\left(-2\right)-3$

$=-7$ .

Therefore, the vertex is at the point $\left(-2,-7\right)$

2. Find the vertex of $y=-x^2-2x+1$ .

Start by calculating the x-value.

$\frac{-b}{2a}$

$\frac{-\left(-2\right)}{2\left(-1\right)}$

$\frac{2}{-2}$

$=-1$

Plug this value into the equation to get the y-value.

$y=-{\left(-1\right)}^2-2\left(-1\right)+1$

$=2$

Therefore, the vertex is $\left(-1,2\right)$ .

3. Find the vertex of $y=-x^2-3x+4$ .

Calculate the x-value.

$\frac{-b}{2a}$

$\frac{-\left(-3\right)}{2\left(-1\right)}$

$\frac{3}{-2}$

$=-\frac{3}{2}$

This is our x-value, so we'll plug it into the equation.

$-{\left(-\frac{3}{2}\right)}^2-3\left(-\frac{3}{2}\right)+4$

$=\frac{25}{4}$

Therefore, our vertex coordinates are $\left(-\frac{3}{2},\frac{25}{4}\right)$ .

4. Find the axis of symmetry of $y=x^2+2x-1$ .

To calculate the axis of symmetry, we use the formula $x=-\frac{b}{2a}$

$-\frac{2}{2\times 1}$

$=-1$

Therefore, the axis of symmetry is $x=-1$ .

5. Find the axis of symmetry of $y=-x^2-4x+1$ .

$\frac{-b}{2a}$

$\frac{-\left(-4\right)}{2\left(-1\right)}$

$=\frac{4}{-2}$

$=-2$

The axis of symmetry is $=-2$ .

6. Find the y-intercept of $y=2x^2+3x+1$ .

To find the y-intercept, all we need to do is plug zero into x.

$y=2{\left(0\right)}^2+3\left(0\right)+1$

$y=1$

Therefore, the y-intercept is at $\left(0,1\right)$ .

7. State the domain and range of $y=x^2+2$ .

Since this spans all possible x-values on a graph, we can say that the domain is $-\infty <x<\infty$ . We can also call this $(-\infty ,\infty )$ in interval notation.

Since the range of y-values begins at 2 and continues upward toward infinity, we can call the domain $2\le y<\infty$ . In interval notation, we would call this $[2,\infty )$ .

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Get additional assistance with quadratic functions

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