What are the x-intercepts of the graph of the function f(x) = x² + 5x – 36?

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To find the x-intercepts of the function f(x) = x² + 5x – 36, we need to determine the points where the graph of the function crosses the x-axis. This occurs when f(x) is equal to zero, so we set the equation to zero:

x² + 5x - 36 = 0

Next, we can solve this quadratic equation using the quadratic formula, which is given by:

x = (-b ± √(b² - 4ac)) / 2a

In our function, a is 1, b is 5, and c is -36. Plugging these values into the quadratic formula:

x = (-(5) ± √((5)² - 4(1)(-36))) / (2(1))

This simplifies to:

x = (-5 ± √(25 + 144)) / 2

Calculating inside the square root:

25 + 144 = 169

Now, taking the square root of 169:

√169 = 13

Now we substitute back into our equation:

x = (-5 ± 13) / 2

This yields two potential solutions:

x₁ = (-5 + 13) / 2 = 8 / 2 = 4
x₂ = (-5 - 13) / 2 = -18 / 2 = -9

Thus, the x-intercepts of the graph of the function f(x) = x² + 5x – 36 are:

  • (4, 0)
  • (-9, 0)

So, these points (4, 0) and (-9, 0) are where the graph meets the x-axis.

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