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Fynjy0 [20]
3 years ago
9

Find the vertex of f(x)=4x^2+19x+12. Please help, thanks!

Mathematics
1 answer:
pickupchik [31]3 years ago
6 0

Answer:

(-19/8, -9/16)

Step-by-step explanation:

Recall that the vertex of a parabola is situated on the axis of symmetry, which here is

      -b         -19

x = ------- = --------- = -2 3/8

      2a           8

To find the y-coordinate of the vertex, evaluate f(x)=4x^2+19x+12 at -19/8.

Using a calculator, we find that f(-19/8) = -10.5625.

Then the vertex is at (-19/8, -10.5625), or (-19/8, -9/16)

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(a^n)^m=a^{nm}\\\\\log_ab=c\iff a^c=b\\\\a^{\log_ax}=x\\\\n\log_ab=\log_ab^n\\\\\log_ab+\log_ac=\log_a(bc)\\============================\\\\1.\\y=3\log_3x^4\to y=\log_3(x^4)^3\to y=\log_3x^{12}

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<span>potential; kinetic is the answer.</span>
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The correct answer is C. 225.

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