Solved on Oct 26, 2023

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

STEP 1

Assumptions1. The function given is $f(x)=-3 x^{}+12 x-8$ . The function is a quadratic function, which is a polynomial of degree3. The standard form of a quadratic function is $f(x)=ax^{}+bx+c$ , where $a$ , $b$ , and $c$ 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 $-\frac{b}{a}$
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 $x=-\frac{b}{a}$

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

The axis of symmetry of the function is a vertical line passing through the vertex. Its equation is $x=-\frac{b}{2a}$ , which we have already calculated as $x=2$ .
So, the function $f(x)=-3 x^{2}+12 x-8$ has a maximum value of $4$ at $x=2$ , and the equation of the axis of symmetry is $x=2$ .

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

STEP 1

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

The axis of symmetry of the function is a vertical line passing through the vertex. Its equation is $x=-\frac{b}{2a}$ , which we have already calculated as $x=2$ .
So, the function $f(x)=-3 x^{2}+12 x-8$ has a maximum value of $4$ at $x=2$ , and the equation of the axis of symmetry is $x=2$ .

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

STEP 1

STEP 2

First, let's identify the values of aaa, bbb, and ccc in the given quadratic function.a=−,b=12,c=−8a = -, 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}−2ab​.x=−b2a=−122(−3)x = -\frac{b}{2a} = -\frac{12}{2(-3)}x=−2ab​=−2(−3)12​

STEP 4

Calculate the x-coordinate of the vertex.x=−12−6=2x = -\frac{12}{-6} =2x=−−612​=2

STEP 5

Now that we have the x-coordinate of the vertex, we can find the y-coordinate by substituting x=2x=2x=2 into the function f(x)f(x)f(x).f(2)=−3(2)2+12(2)−8f(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)+24−8=−12+24−8=4f(2) = -3(4)+24-8 = -12+24-8 =4f(2)=−3(4)+24−8=−12+24−8=4

STEP 7

The vertex of the function is (2,4)(2,4)(2,4). Since a=−3a=-3a=−3 is negative, the function opens downwards, meaning the vertex is a maximum point. So, the maximum value of the function is 444.

STEP 8

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

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

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

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

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

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

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