Analyze Domain Behavior: Finding Increasing and Decreasing Intervals of f(x)=5x²-25

Q: How do I remember which interval is increasing vs decreasing?

Think of the parabola shape! Since $ a = 5 > 0 $, it's a U-shape. Going left from the vertex (x going down the left side = decreasing. Going right (x > 0) means going up the right side = increasing.

Q: What if the coefficient of x² was negative?

If \( a , the parabola would flip upside down (∩-shape)! Then it would be increasing for x decreasing for x > 0 - exactly opposite of our current function.

Q: Do I need to find the actual vertex point?

For increasing/decreasing intervals, you only need the x-coordinate of the vertex. The y-coordinate $ f(0) = -25 $ tells you the minimum value but doesn't affect the intervals.

Q: Can I use a graph to check my answer?

Absolutely! Graphing $ f(x) = 5x^2 - 25 $ shows a U-shaped parabola with its lowest point at $ (0, -25) $. You can visually confirm the function decreases left of x = 0 and increases right of x = 0.

Quadratic Functions with Vertex Analysis

What are the the increasing and decreasing domains of the function below?

$f(x)=5x^2-25$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the domain of decrease and increase of the function

00:03 Notice the coefficient of X squared is positive, therefore the function is happy

00:07 Let's look at the coefficients of the trinomial

00:11 We'll use the formula to find the vertex of the parabola

00:16 We'll substitute appropriate values according to the given data and solve to find the vertex

00:21 This is the X value at the vertex of the parabola

00:24 We'll determine when the parabola decreases and increases based on its type

00:31 We'll draw the X-axis and find the domain of decrease and increase

00:43 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

What are the the increasing and decreasing domains of the function below?

$f(x)=5x^2-25$

Step-by-step solution

To determine the increasing and decreasing domains of the quadratic function $f(x) = 5x^2 - 25$ , we begin by analyzing its structure:

This function is a quadratic function of the form $ax^2 + c$ . Here, $a = 5$ , which is positive. As such, the parabola opens upwards.

The vertex of such a quadratic function, when $b = 0$ , is simply at $x = 0$ . Thus, the symmetry point of the parabola is based on this vertex.

Since the parabola opens upwards:

The function is decreasing on the interval $(-\infty, 0)$ .
The function is increasing on the interval $(0, \infty)$ .

Therefore, the function $f(x) = 5x^2 - 25$ is:

$x < 0$ decreasing

$0 < x$ increasing

Thus, the correct answer choice for the intervals is the one provided in Choice 4.

Final Answer

$x < 0$ decreasing

$0 < x$ increasing

Key Points to Remember

Essential concepts to master this topic

Vertex Rule: For $f(x) = ax^2 + c$ , vertex is at $x = 0$
Sign Analysis: When $a = 5 > 0$ , parabola opens upward from vertex
Interval Check: Test values like $f(-1) = 5(-1)^2 - 25 = -20$ and $f(1) = -20$ ✓

Common Mistakes

Avoid these frequent errors

Confusing the vertex location with coefficient values
Don't think the vertex is at $x = 5$ just because you see the number 5 in the function = wrong intervals! The constant term -25 doesn't affect the vertex position. Always identify the vertex by the structure: for $f(x) = ax^2 + c$ , the vertex is always at $x = 0$ .

Practice Quiz

Test your knowledge with interactive questions

Find the ascending area of the function

$$ f(x)=2x^2 $$

FAQ

Everything you need to know about this question

Why is the vertex at x = 0 and not x = 5?

Great question! The vertex occurs where the derivative equals zero. For $f(x) = 5x^2 - 25$ , we get $f'(x) = 10x = 0$ , so x = 0. The numbers 5 and -25 are just coefficients, not the vertex location.

How do I remember which interval is increasing vs decreasing?

Think of the parabola shape! Since $a = 5 > 0$ , it's a U-shape. Going left from the vertex (x < 0) means going down the left side = decreasing. Going right (x > 0) means going up the right side = increasing.

What if the coefficient of x² was negative?

If $a < 0$ , the parabola would flip upside down (∩-shape)! Then it would be increasing for x < 0 and decreasing for x > 0 - exactly opposite of our current function.

Do I need to find the actual vertex point?

For increasing/decreasing intervals, you only need the x-coordinate of the vertex. The y-coordinate $f(0) = -25$ tells you the minimum value but doesn't affect the intervals.

Can I use a graph to check my answer?

Absolutely! Graphing $f(x) = 5x^2 - 25$ shows a U-shaped parabola with its lowest point at $(0, -25)$ . You can visually confirm the function decreases left of x = 0 and increases right of x = 0.

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