Find the range of f(x) = 2x^2 + 6x – 8.

Find the range of f(x) = 2x^2 + 6x – 8. (Enter your answer using interval notation

The Correct Answer and Explanation is:

To find the range of the quadratic function f(x)=2×2+6x−8f(x) = 2x^2 + 6x – 8f(x)=2×2+6x−8, we can follow a few key steps:

Step 1: Identify the shape of the graph

The function is a quadratic equation, and since the coefficient of x2x^2×2 (which is 2) is positive, the graph is a parabola that opens upward.

Step 2: Find the vertex

The range of a parabola depends on its vertex. The vertex of a parabola represented by f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c can be found using the formula for the x-coordinate of the vertex:xvertex=−b2ax_{\text{vertex}} = \frac{-b}{2a}xvertex​=2a−b​

For the function f(x)=2×2+6x−8f(x) = 2x^2 + 6x – 8f(x)=2×2+6x−8:

  • a=2a = 2a=2
  • b=6b = 6b=6

Thus, the x-coordinate of the vertex is:xvertex=−62(2)=−64=−32x_{\text{vertex}} = \frac{-6}{2(2)} = \frac{-6}{4} = -\frac{3}{2}xvertex​=2(2)−6​=4−6​=−23​

Step 3: Find the y-coordinate of the vertex

Substitute x=−32x = -\frac{3}{2}x=−23​ back into the original function to find the y-coordinate of the vertex.f(−32)=2(−32)2+6(−32)−8f\left(-\frac{3}{2}\right) = 2\left(-\frac{3}{2}\right)^2 + 6\left(-\frac{3}{2}\right) – 8f(−23​)=2(−23​)2+6(−23​)−8

First, square −32-\frac{3}{2}−23​:(−32)2=94\left(-\frac{3}{2}\right)^2 = \frac{9}{4}(−23​)2=49​

Now substitute into the function:f(−32)=2×94+6×(−32)−8f\left(-\frac{3}{2}\right) = 2 \times \frac{9}{4} + 6 \times \left(-\frac{3}{2}\right) – 8f(−23​)=2×49​+6×(−23​)−8f(−32)=184−9−8f\left(-\frac{3}{2}\right) = \frac{18}{4} – 9 – 8f(−23​)=418​−9−8f(−32)=184−17=184−684=−504=−12.5f\left(-\frac{3}{2}\right) = \frac{18}{4} – 17 = \frac{18}{4} – \frac{68}{4} = \frac{-50}{4} = -12.5f(−23​)=418​−17=418​−468​=4−50​=−12.5

So, the vertex is at (−32,−12.5)\left(-\frac{3}{2}, -12.5\right)(−23​,−12.5).

Step 4: Determine the range

Since the parabola opens upward, the lowest value of the function is the y-coordinate of the vertex, −12.5-12.5−12.5, and the function increases without bound as xxx moves away from the vertex. Therefore, the range of the function is:Range=[−12.5,∞)\text{Range} = [-12.5, \infty)Range=[−12.5,∞)

Final Answer:

The range of f(x)=2×2+6x−8f(x) = 2x^2 + 6x – 8f(x)=2×2+6x−8 is:[−12.5,∞)[-12.5, \infty)[−12.5,∞)

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