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Find the vertex of the following quadratic function:𝑓(𝑥)=𝑥2+3𝑥−5Group of answer choices(4,3)(−714,−112)(4,−5)(−112,−714)

Question

Find the vertex of the following quadratic function:

f(x)=x2+3x5 f(x) = x^2 + 3x - 5

Group of answer choices:

  • (4, 3)
  • (−714, −112)
  • (4, −5)
  • (−112, −714)
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Solution

The vertex of a quadratic function given in the form f(x) = ax^2 + bx + c is given by the point (-b/2a, f(-b/2a)).

Here, a = 1, b = 3, and c = -5.

Step 1: Calculate -b/2a = -3/(2*1) = -3/2 = -1.5

Step 2: Substitute x = -1.5 into the function to find the y-coordinate of the vertex:

f(-1.5) = (-1.5)^2 + 3*(-1.5) - 5 = 2.25 - 4.5 - 5 = -7.25

So, the vertex of the function f(x) = x^2 + 3x - 5 is (-1.5, -7.25).

None of the given options match this result.

This problem has been solved

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