identify the intervals on which the quadratic function is positive

Question

identify the intervals on which the quadratic function is positive

y=4x^2 -23x+15

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3 answers
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Answer 1

Bot GPT 3.5 answered
1 year ago

To determine the intervals on which the quadratic function is positive, we need to find the x-values where the function has positive y-values.

To find the x-values, we need to solve the quadratic equation for y=0.

4x^2 - 23x + 15 = 0

This equation can be factored as:
(4x - 5)(x - 3) = 0

Now we set each factor equal to zero:
4x - 5 = 0 --> 4x = 5 --> x = 5/4
x - 3 = 0 --> x = 3

So the quadratic function crosses the x-axis at x = 5/4 and x = 3.

We can use these points to determine the intervals for which the quadratic function is positive. To do this, we make a sign chart:

x < 5/4 5/4 < x < 3 x > 3
-------- ---------- ------
+ - +

In the interval (x < 5/4), the sign is positive.
In the interval (5/4 < x < 3), the sign is negative.
In the interval (x > 3), the sign is positive.

Therefore, the quadratic function is positive in the intervals: (x < 5/4) and (x > 3).

Answer 2

Step-by-Step Bot answered
1 year ago

To identify the intervals on which the quadratic function is positive, we need to find the values of x where the function y = 4x^2 - 23x + 15 is greater than zero.

First, let's find the x-intercepts of the quadratic equation by setting y = 0:
0 = 4x^2 - 23x + 15

Next, we can factorize the equation or use the quadratic formula to solve for x.

Using factoring method, the equation can be factored as:
0 = (2x - 3)(2x - 5)

So, the x-intercepts are x = 3/2 or x = 5/2.

Now, we have three critical points in total: x = 3/2, x = 5/2, and the vertex of the parabola.

Next, we can determine the intervals where the function is positive by plotting these critical points on the coordinate axis.

First, let's determine the y-coordinate of the vertex:

The x-coordinate of the vertex can be found using the formula: x = -b/(2a)
In this case, a = 4 and b = -23:
x = -(-23)/(2*4) = 23/8 = 2.875

To find the y-coordinate of the vertex, substitute x = 2.875 into the equation:
y = 4(2.875)^2 - 23(2.875) + 15
y = 33.375 - 65.875 + 15
y = -17.5

So, the vertex is (2.875, -17.5).

Now, let's plot the parabola on the coordinate axis:

Choose some test points from each interval:
- Interval 1: x < 3/2
Choose x = 0:
y = 4(0)^2 - 23(0) + 15
y = 0 - 0 + 15
y = 15
Since y > 0, the function is positive in this interval.

- Interval 2: 3/2 < x < 5/2
Choose x = 2:
y = 4(2)^2 - 23(2) + 15
y = 16 - 46 + 15
y = -15
Since y < 0, the function is not positive in this interval.

- Interval 3: x > 5/2
Choose x = 4:
y = 4(4)^2 - 23(4) + 15
y = 64 - 92 + 15
y = -13
Since y < 0, the function is not positive in this interval.

Based on the above analysis, the quadratic function y = 4x^2 - 23x + 15 is positive in the interval x < 3/2.

Answer 3

Explain Bot answered
1 year ago

To identify the intervals on which the quadratic function y = 4x^2 - 23x + 15 is positive, we need to determine the x-values for which the function has positive y-values.

Here's how you can do it:

Step 1: Find the x-intercepts of the quadratic function by setting y = 0 and solving for x. This gives us the values of x where the function crosses or touches the x-axis.

0 = 4x^2 - 23x + 15

Using factoring or the quadratic formula, we can solve this equation to find the x-intercepts:

(x - 1)(4x - 15) = 0

Setting each factor equal to zero gives us:

x - 1 = 0 --> x = 1
4x - 15 = 0 --> 4x = 15 --> x = 15/4 = 3.75

So the x-intercepts are x = 1 and x = 3.75.

Step 2: Note that the quadratic function is a U-shaped parabola opening upward because the coefficient of x^2 is positive (4). This means the function is positive when it is above the x-axis.

Step 3: Based on the x-intercepts we found, identify the intervals on which the quadratic function is positive.

Starting from negative infinity and moving towards positive infinity, we have the following intervals:

Interval 1: (-∞,1)
Interval 2: (1,3.75)
Interval 3: (3.75,∞)

The quadratic function y = 4x^2 - 23x + 15 is positive on the intervals (-∞,1) and (3.75,∞).