finding the range of a quadratic function?

Question

How would you find the range of this quadratic function? Should you rewrite it in a different form? Regardless of how , can someone show steps so I can see how to do it?
y=x^2 +8x - 2

Answer 1

y=(x+4)^2 - 18

range = [-18, inf)

Answer 2

The "range" is all the possible values the function can take, while the "domain" is all the values that the variable in the function can take.

Usually, it's easiest if you can convert it to parabola form

y = x^2 + 8x - 2

==> y = x^2 + 8x + 16 - 16 -2

==>y = (x+4)^2 - 18

So, you know that the lowest value for y occurs when x = -4, and that is y = -18



You can figure out the rest...

Answer 3

Complete the square and rewrite the equation as: y = (x+4)^2 - 18. The minimum y-value is -18, when x = -4. The range is y is greater than or equal to -18.

Answer 4

Convert to vertex form, y = a(x - h)² + k, where (h, k) is the vertex:



y = x² + 8x - 2

y = (x² + 8x) - 2

y = (x² + 8x + 16) - 2 - 16

y = (x + 4)² - 18



Vertex (- 4, - 18)



Since the equation is y as a function of x, the parabola opens vertically and, since the exponent of the squared term is positive, it opens upward, so



Range: - 18 ≤ y < ∞, or y ≥ - 18

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