Determine the maximum or minimum value. Use at least two different methods.
y=3x(x-2)+5
One method to use would be completng the square.
3x^2 -6x +5 = 3(x^2 -2x +1) +2
= 3(x-1)^2 +2
The lowest possible value is clearly 2. It happens when x = 1. There is no maximum value.
Another method would be calculus. A relative maximum or minimum occurs where dy/dx = 0
6x -6 = 0
x = 1
y = 2.
Given:Y = 3X(X - 2) + 5
Remove parenthesis:
Y = 3X^2 -6X + 5
Find derivitive (Y') and set to 0:
Y' = 6X -6 = 0, X = 1 = h = X-coordinate of vertex.
Y = 3(1)(1 - 2) + 5 = 2 = MIN. = K
= Y coordinate of vertex. V(h,k)
V(1,2).
2nd method
X =-B/2A = 6/6 = 1 = h = X-coordinate
of vertex.
Y = 3(1)(1 -2) + 5 = 2 = min. = k = Y-
coordinate of vertex. V(h,k) = V(1,2).
Since the parabola opens upward, Y = 2
is a minimum.
To determine the maximum or minimum value of the function y = 3x(x - 2) + 5, we can use two different methods: completing the square and finding the vertex of the quadratic function.
Method 1: Completing the Square
To complete the square, we'll rewrite the quadratic equation in vertex form.
Step 1: Expand the equation:
y = 3x(x - 2) + 5
y = 3x^2 - 6x + 5
Step 2: Group the quadratic terms together:
y = (3x^2 - 6x) + 5
Step 3: Factor out the common coefficient of x^2 and x:
y = 3(x^2 - 2x) + 5
Step 4: Complete the square by adding and subtracting the square of half the coefficient of x:
y = 3(x^2 - 2x + (-2/2)^2) + 5 - 3(-2/2)^2
y = 3(x^2 - 2x + 1) + 5 - 3(1)
y = 3(x - 1)^2 + 5 - 3
y = 3(x - 1)^2 + 2
Step 5: From the equation, we can see that the vertex of the parabola is (1, 2). Since the coefficient of x^2 is positive, the parabola opens upwards. Therefore, the vertex represents the minimum value of the function.
So, the minimum value is y = 2.
Method 2: Finding the Vertex
Alternatively, we can find the vertex by using the formula x = -b/2a.
Step 1: Identify the quadratic equation in standard form: y = ax^2 + bx + c.
In our case, a = 3, b = -6, and c = 5.
Step 2: Find the x-coordinate of the vertex: x = -(-6) / (2 * 3) = 1.
Step 3: Substitute the value of x into the equation to find the y-coordinate of the vertex: y = 3(1)^2 - 6(1) + 5 = 2.
Again, we find that the minimum value is y = 2 at the vertex (1, 2).
In conclusion, the minimum value of the function y = 3x(x - 2) + 5 is 2.
To determine the maximum or minimum value of the given function, y = 3x(x - 2) + 5, we can use two methods: completing the square and finding the vertex using the formula for x-coordinate of the vertex.
Method 1: Completing the Square
Step 1: Rearrange the equation to the standard form, y = ax^2 + bx + c.
y = 3x(x - 2) + 5
y = 3x^2 - 6x + 5
Step 2: Divide the coefficient of x by 2 and square the result to determine the value to add and subtract inside the parentheses. In this case, it is (-6/2)^2 = 9.
Step 3: Add and subtract the value obtained in step 2 inside the parentheses.
y = 3x^2 - 6x + 9 - 9 + 5
y = 3x^2 - 6x + 9 - 4
Step 4: Group the terms and factorize the perfect square trinomial.
y = [3x^2 - 6x + 9] - 4
y = 3(x^2 - 2x + 3) - 4
y = 3(x - 1)^2 - 4
Step 5: Since the expression (x - 1)^2 is always nonnegative, the minimum value occurs when it equals zero. Thus, the minimum value is -4, which occurs when x - 1 = 0, or x = 1.
Therefore, the minimum value of the given function is -4.
Method 2: Finding the Vertex using the Formula
Step 1: The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
Step 2: Identify the coefficients of the equation in standard form:
a = 3
b = -6
Step 3: Substituting the values into the formula, we get:
x = -(-6) / (2 * 3)
x = 6 / 6
x = 1
Step 4: To find the corresponding y-coordinate, substitute the value of x into the original equation:
y = 3x(x - 2) + 5
y = 3(1)(1 - 2) + 5
y = 3(1)(-1) + 5
y = -3 + 5
y = 2
Therefore, the vertex of the parabola is (1, 2), and the minimum value is 2.
In conclusion, the function y = 3x(x - 2) + 5 has a minimum value of -4 when using the completing the square method and a minimum value of 2 when using the vertex formula to find the vertex.