What are the x-intercepts of the function f(x) = 2x^2 + 3x – 20?

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To find the x-intercepts of the function f(x) = 2x2 + 3x – 20, we need to determine the points where the graph of the function intersects the x-axis. This occurs when the value of the function is equal to zero.

So, we set the function equal to zero:

2x2 + 3x - 20 = 0

This is a quadratic equation in the form of ax2 + bx + c = 0, where:

  • a = 2
  • b = 3
  • c = -20

To solve for x, we can use the quadratic formula:

x = (-b ± √(b2 - 4ac)) / 2a

Plugging in our values:

x = (-(3) ± √((3)2 - 4(2)(-20))) / (2(2))

Now, calculate the discriminant:

b2 - 4ac = 32 - 4(2)(-20) = 9 + 160 = 169

Now, we can substitute back to find the x-intercepts:

x = (-3 ± √169) / 4

This simplifies to:

x = (-3 ± 13) / 4

We will now calculate both possible values for x:

  1. x = (-3 + 13) / 4 = 10 / 4 = 2.5
  2. x = (-3 – 13) / 4 = -16 / 4 = -4

Thus, the x-intercepts of the function are:

  • (2.5, 0)
  • (-4, 0)

In conclusion, the x-intercepts of the function f(x) = 2x2 + 3x – 20 are (2.5, 0) and (-4, 0).

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