What's the source for this problem? I can see three relatively straightforward ways to solve it (verifying real roots, calculus, or the vertex formula for parabolas), but I haven't encountered questions that use these techniques on the GRE before.
I'm curious to see if anyone has an alternate solution.
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Plugging in and verifying that we have real roots Start with the lowest value for f(x) and plug in.
f(x) = x^2 + 4x - 5
-9 = x^2 + 4x - 5
Set the quadratic equation equal to zero to factor:
0 = x^2 + 4x + 4
0 = (x+2)(x+2)
x = -2
Since we have non-imaginary roots, -9 is a reasonable value for f(x), and it was the smallest of our answer choices.
We can find the x-coordinate of the minimum or maximum of a function by taking the derivative and setting it equal to 0.
f(x) = x^2 + 4x - 5
f'(x) = 2x + 4
0 = 2x + 4
-4 = 2x
-2 = x
We can then plug in -2 as our x value and solve for f(x)
f(x) = (-2)^2 + 4(-2) - 5
f(x) = 4 - 8 - 5
f(x) = -9
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Using the parabola vertex formula Given a parabola in the form f(x) = ax^2 +bx +c, we can identify the x-coordinate of the vertex using this formula:
vertex x-coordinate = -b / 2a
vertex x-coordinate = -4 / 2(1)
vertex x-coordinate = -2
We can then plug that x value into the equation and get the same result:
f(x) = (-2)^2 + 4(-2) - 5
f(x) = 4 - 8 - 5
f(x) = -9
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