Find the range of the quadratic function.

Question

y = 2x² + 8x + 9

Write your answer as an inequality using

or x or y as appropriate.

Or, you may instead click on "Empty set" or "All reals" as the answer.

Answer 1

There is a small error in the other tutor's solution. The range is the set of y values; the domain is the set of all x values.

The smallest y value is 1, not -2. So, in interval notation, the range is [1,∞).

The problem actually asks for the range in inequality notation, so:

y ≥ 1

is the range in this notation.

The domain by the way, in this case, is all real numbers.

Answer 2

The graph of the quadratic function is an upward opening parabola (because the numerical coefficient of the squared term is positive).

If you could find the coordinates of the vertex, then the range would be from the y-coordinate of the vertex through positive infinity.

There are a few ways to find the vertex, probably the easiest in this case is to understand that the axis of symmetry is given by the equation x = -b/2a where a = 2, b = 8 (and c = 9). Since the vertex lies on the axis of symmetry (which is x = -8/4) the vertex has an x-coordinate of -2. To find the y-coordinate of the vertex substitute x = -2 into the function definition.

y = 2(-2)² + 8(-2) + 9

y = 2(4) - 16 + 9

y = 1

So the vertex is located at (-2,1), which is the low point on the parabola.

That means the range is [1, ∞).

It is not possible to get a y-value outside that interval no matter the x-value.

desmos.com/calculator/nhsinh8lld