Solved on Oct 26, 2023

Find the min/max value and axis of symmetry for f(x)=3x2+12x8f(x)=-3 x^{2}+12 x-8. The yy-value of the extremum is the solution.

STEP 1

Assumptions1. The function given is f(x)=3x+12x8f(x)=-3 x^{}+12 x-8 . The function is a quadratic function, which is a polynomial of degree3. The standard form of a quadratic function is f(x)=ax+bx+cf(x)=ax^{}+bx+c, where aa, bb, and cc are constants4. The maximum or minimum value of a quadratic function occurs at its vertex5. The x-coordinate of the vertex of a quadratic function is given by ba-\frac{b}{a}
6. The y-coordinate of the vertex of a quadratic function is found by substituting the x-coordinate of the vertex into the function7. The axis of symmetry of a quadratic function is a vertical line passing through the vertex of the function, and its equation is x=bax=-\frac{b}{a}

STEP 2

First, let's identify the values of aa, bb, and cc in the given quadratic function.
a=,b=12,c=8a = -, b =12, c = -8

STEP 3

Next, we calculate the x-coordinate of the vertex using the formula b2a-\frac{b}{2a}.
x=b2a=122(3)x = -\frac{b}{2a} = -\frac{12}{2(-3)}

STEP 4

Calculate the x-coordinate of the vertex.
x=126=2x = -\frac{12}{-6} =2

STEP 5

Now that we have the x-coordinate of the vertex, we can find the y-coordinate by substituting x=2x=2 into the function f(x)f(x).
f(2)=3(2)2+12(2)8f(2) = -3(2)^{2}+12(2)-8

STEP 6

Calculate the y-coordinate of the vertex.
f(2)=3(4)+248=12+248=4f(2) = -3(4)+24-8 = -12+24-8 =4

STEP 7

The vertex of the function is (2,4)(2,4). Since a=3a=-3 is negative, the function opens downwards, meaning the vertex is a maximum point. So, the maximum value of the function is 44.

STEP 8

The axis of symmetry of the function is a vertical line passing through the vertex. Its equation is x=b2ax=-\frac{b}{2a}, which we have already calculated as x=2x=2.
So, the function f(x)=3x2+12x8f(x)=-3 x^{2}+12 x-8 has a maximum value of 44 at x=2x=2, and the equation of the axis of symmetry is x=2x=2.

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