Determine the X Values: Inequality in Quadratics with y = -3/5x²

Q: Why isn't the answer 'all x' if the parabola opens downward?

Great question! While the parabola does open downward, it touches the x-axis at x = 0. Since we need $ f(x) (strictly less than), we must exclude the point where \( f(x) = 0 $.

Q: How do I know when a parabola opens up or down?

Look at the coefficient of $ x^2 $! If it's positive, the parabola opens upward (U-shape). If it's negative like $ -\frac{3}{5} $, it opens downward (∩-shape).

Q: What does 'x ≠ 0' mean exactly?

It means all real numbers except zero. So x can be any positive or negative number, including fractions and decimals, but not zero itself.

Q: Can I solve this by factoring?

You could factor out the coefficient: $ -\frac{3}{5}x^2 = -\frac{3}{5} \cdot x^2 $, but since there's no linear term, analyzing the sign directly is much faster for this type of problem.

Q: Why is f(0) = 0 important here?

Because the inequality asks for $ f(x) (strictly less than zero). Since \( f(0) = 0 $, we must exclude x = 0 from our solution set.

Quadratic Inequalities with Negative Coefficients

Given the function $y=-\frac{3}{5}x^2$

Determine for which values of x the following holds:

$f(x) < 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Given the function $y=-\frac{3}{5}x^2$

Determine for which values of x the following holds:

$f(x) < 0$

Step-by-step solution

To solve the problem of finding when $f(x) = -\frac{3}{5}x^2 < 0$ , we utilize properties of quadratic functions:

The given function is $y = -\frac{3}{5}x^2$ , where the coefficient of $x^2$ is $-\frac{3}{5}$ .
The negative coefficient indicates the parabola opens downwards.
For any quadratic function of the form $y = ax^2$ where $a < 0$ , the function is negative for all $x$ except at $x = 0$ , where it equals zero.

Therefore, the function $f(x) = -\frac{3}{5}x^2$ is negative for all $x$ except $x = 0$ .

Thus, the set of $x$ satisfying the condition $f(x) < 0$ is $x \neq 0$ .

Hence, the solution is $x \ne 0$ .

Final Answer

$x\ne0$

Key Points to Remember

Essential concepts to master this topic

Parabola Direction: Negative coefficient $a < 0$ means downward opening
Zero Point: Function equals zero only at $x = 0$ since $-\frac{3}{5}(0)^2 = 0$
Check: Test any non-zero value: $-\frac{3}{5}(1)^2 = -\frac{3}{5} < 0$ ✓

Common Mistakes

Avoid these frequent errors

Thinking the function is positive for some x values
Don't assume a quadratic can be positive when the coefficient is negative = wrong solution set! Since $-\frac{3}{5} < 0$ and $x^2 ≥ 0$ always, their product is always ≤ 0. Always remember that negative times non-negative equals non-positive.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of $$ x $$ where $f\left(x\right) > 0$ .

FAQ

Everything you need to know about this question

Why isn't the answer 'all x' if the parabola opens downward?

Great question! While the parabola does open downward, it touches the x-axis at x = 0. Since we need $f(x) < 0$ (strictly less than), we must exclude the point where $f(x) = 0$ .

How do I know when a parabola opens up or down?

Look at the coefficient of $x^2$ ! If it's positive, the parabola opens upward (U-shape). If it's negative like $-\frac{3}{5}$ , it opens downward (∩-shape).

What does 'x ≠ 0' mean exactly?

It means all real numbers except zero. So x can be any positive or negative number, including fractions and decimals, but not zero itself.

Can I solve this by factoring?

You could factor out the coefficient: $-\frac{3}{5}x^2 = -\frac{3}{5} \cdot x^2$ , but since there's no linear term, analyzing the sign directly is much faster for this type of problem.

Why is f(0) = 0 important here?

Because the inequality asks for $f(x) < 0$ (strictly less than zero). Since $f(0) = 0$ , we must exclude x = 0 from our solution set.

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