# How to Find Domain and Range of a Quadratic Function

The domain of a quadratic function in standard form is always all real numbers, meaning you can substitute any real number for *x*. The range of a function is the set of all real values of *y* that you can get by plugging real numbers into *x*.

## Domain and Range of Quadratic Functions

Find the domain and range of y equals 2x squared plus 8x plus 11. **The domain of a quadratic function of this form which is in standard form is always all real numbers.** We can start with that; the domain is all real numbers. What that means is we can use any real value for x. We can substitute any real number for our x. The range for quadratic function, to find that we first have to find the vertex of our quadratic function. To find the vertex of our quadratic function, first we find the x value or the x coordinate of our vertex. To find the x coordinates our vertex we use the formula x equals the opposite of b divided by 2 times a. Not only will this find you the x coordinate of the vertex.

It’s also how you find the axis of symmetry that imaginary vertical line that cuts right down the center of your parabola. These letters a and b come from the standard form of a quadratic function. The standard form of a quadratic function is y equals ax squared plus bx plus c. Our a, b, and c are these values here. To find the x coordinate all we need is our a and our b. x equals the opposite b, b is 8, the opposite of 8 is negative 8 divided by 2 times a, and are a is 2, it’s the leading coefficient 2 times 2. Which is equal to negative eight divided by 2 times 2 is 4 and negative 8 divided by 4 is negative 2. The x value or the x coordinate of our vertex is negative 2. What we’re really focused on is the y coordinate since range is our set of y values. To find the y coordinate of our vertex we take our x value, negative 2, and substitute it into our equation to find y.

y equals 2 times, again our x is negative 2, we’re substituting negative 2 for x squared plus 8 times, again we’re substituting negative 2 for x, plus 11. **Now when we simplify this, we’ll have found our y value.** y again is equal to 2 times, and according to PEMDAS, which is our order of operations, we don’t have any grouping symbols that’s parentheses, but we do have exponent so that’s where we’re going to start. Negative 2 squared is positive 4, a negative times a negative as a positive. I’m going to go ahead and simplify this here there’s no reason why not to. 8 times negative 2 is negative 16, so minus 16 plus 11. y equals 2 times 4 is 8, minus 16, plus 11. Now that we just have addition and subtraction, we just work it from left to right. y equals 8 minus 16 is negative 8, plus 11. y equals negative 8 plus 11, which is a positive 3.

The x value of our vertex is negative 2 and the y value is 3, which means our vertex is negative 2, 3. What does this mean for finding our range? Well, since this is a quadratic function that means when it’s graft it’s going to be a parabola, and knowing what our vertex is I’ll just draw a little rough drawing here to help us visualize, our vertex is at negative 2, 3. Negative 1, 2, up 1, 2, 3. Since our a our leading coefficient is positive, that means our parabola is concave up. Well, by finding our vertex we found the lowest point on our quadratic function, we found the minimum value, the minimum y value.

If you look at it now you can see how our domain would be all real numbers, because left to right this is going to keep going up and out as it goes and eventually it’s going to hit every single x value, which is why our domain is all real numbers. Our range in this case is the lowest y value the lowest this parabola goes is right here down to our vertex is a positive 3. It’s all numbers greater than or equal to 3 because it’s 3 and then it goes up. All real numbers greater than or equal to 3. If this parabola had been concave down instead of concave up, say this is y equals negative 2x squared instead of a positive 2x squared, it was concave down instead then our answer would change just a little bit in that instead of it being all numbers that are greater than or equal to 3, the vertex would be our maximum point of the parabola it would be the highest point, it’d be all real numbers less than or equal to positive 3. That’s how you find a domain and range for quadratic functions.