identify the intervals of y=5x^2-41x+42
To identify the intervals of the function y = 5x^2 - 41x + 42, we need to determine where the function is increasing or decreasing.
To do this, we can take the derivative of the function, which will give us the slope of the tangent line at any given point.
The derivative of y = 5x^2 - 41x + 42 is y' = 10x - 41.
Setting y' equal to zero and solving for x, we get:
10x - 41 = 0
10x = 41
x = 41/10
This gives us the critical point of the function.
To determine the intervals of increasing and decreasing, we can use test values within each interval.
For values of x less than 41/10 (approximately 4.1), if we plug them into y' = 10x - 41, the result will be negative. This means that the function is decreasing in this interval.
For values of x greater than 41/10, if we plug them into y' = 10x - 41, the result will be positive. This means that the function is increasing in this interval.
Therefore, the function y = 5x^2 - 41x + 42 is decreasing when x is less than 41/10, and it is increasing when x is greater than 41/10.
So the intervals of the function are:
(-infinity, 41/10) for decreasing values of y
(41/10, +infinity) for increasing values of y
find the interval(s) of y=5x^2-41x+42
To find the intervals of the function y = 5x^2 - 41x + 42, we need to determine where the function is increasing or decreasing.
First, let's find the critical points of the function by finding the x-values where the derivative of the function is equal to zero or undefined.
Finding the derivative of y = 5x^2 - 41x + 42:
y' = 10x - 41
Setting y' equal to zero:
10x - 41 = 0
10x = 41
x = 4.1
Since the derivative is always defined for every x-value, we only have one critical point at x = 4.1.
Now, we can determine the intervals of increase and decrease by testing points within each interval.
For x values less than 4.1, let's test x = 0:
y' = 10(0) - 41 = -41
Since the derivative is negative, the function is decreasing in this interval.
For x values greater than 4.1, let's test x = 5:
y' = 10(5) - 41 = 9
Since the derivative is positive, the function is increasing in this interval.
Therefore, we can conclude that the function y = 5x^2 - 41x + 42 is decreasing for x < 4.1 and increasing for x > 4.1.
Thus, the interval(s) for y = 5x^2 - 41x + 42 are:
(-infinity, 4.1) for the interval of decreasing values of y
(4.1, +infinity) for the interval of increasing values of y
To identify the intervals of the quadratic function y = 5x^2 - 41x + 42, we need to find the values of x where the function is either increasing or decreasing.
First, let's find the x-coordinate of the vertex, which is the turning point of the parabola. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a, b, and c are the coefficients of the quadratic function in the form ax^2 + bx + c.
For the given quadratic function y = 5x^2 - 41x + 42, a = 5 and b = -41.
x = -( -41 ) / (2 * 5)
x = 41 / 10
x = 4.1
So, the x-coordinate of the vertex is 4.1.
Now, let's examine the value of the coefficient of x^2, which is a = 5. Since the coefficient is positive, the parabola opens upward.
Since the parabola is upward-facing and the coefficient of x^2 is positive, the quadratic function is decreasing to the left of the vertex and increasing to the right of the center.
Therefore, the interval where the function is decreasing is (-∞, 4.1) and the interval where the function is increasing is (4.1, +∞).
To identify the intervals of the given quadratic function y = 5x^2 - 41x + 42, we need to find the values of x where the function is positive or negative.
The function is in the form of a quadratic equation, which will result in a parabolic shape. To determine the intervals, we will find the x-values where the graph of the quadratic crosses or touches the x-axis.
Step 1: Find the x-intercepts
To find the x-intercepts, we set y = 0 and solve for x. So, we have:
0 = 5x^2 - 41x + 42
This equation can be factored or solved using the quadratic formula. Let's solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 5, b = -41, and c = 42. Plugging these values into the formula, we get:
x = (-(-41) ± √((-41)^2 - 4 * 5 * 42)) / (2 * 5)
x = (41 ± √(1681 - 840)) / 10
x = (41 ± √841) / 10
x = (41 ± 29) / 10
So, we have two x-intercepts:
x1 = (41 + 29) / 10 = 7
x2 = (41 - 29) / 10 = 1.2
Step 2: Determine the intervals
Now that we have the x-intercepts, we can determine the intervals of the function.
We arrange the x-intercepts in increasing order: x2, x1. So, the intervals are:
Interval 1: (-∞, x2) (-∞, 1.2)
Interval 2: (x2, x1) (1.2, 7)
Interval 3: (x1, +∞) (7, +∞)
To check if the function is positive or negative within these intervals, we can select a test point from each interval and substitute it into the equation. If the value is positive, then the function is positive in that interval. If the value is negative, then the function is negative.
Let's select the following test points:
- Any number less than x2 (e.g., x = 0)
- Any number between x2 and x1 (e.g., x = 3)
- Any number greater than x1 (e.g., x = 10)
Step 3: Substitute the test points
For x = 0:
y = 5(0)^2 - 41(0) + 42 = 42
Since the value is positive, the function is positive in Interval 1: (-∞, 1.2)
For x = 3:
y = 5(3)^2 - 41(3) + 42 = 12
Since the value is positive, the function is positive in Interval 2: (1.2, 7)
For x = 10:
y = 5(10)^2 - 41(10) + 42 = -208
Since the value is negative, the function is negative in Interval 3: (7, +∞)
Therefore, the intervals of y = 5x^2 - 41x + 42 are:
Interval 1: (-∞, 1.2)
Interval 2: (1.2, 7)
Interval 3: (7, +∞)