Solve y=-(1/3x+15)²: Finding Values Where Function is Negative

Look at the function below:

$y=-\left(\frac{1}{3}x+15\right)^2$

Then determine for which values of $x$ the following is true:

$f(x) < 0$

To solve this problem, let's examine the function $y = -\left(\frac{1}{3}x + 15\right)^2$ and determine when $y < 0$.

Step 1: Analyze the expression inside the function.
The function involves the square of a linear term, $\left(\frac{1}{3}x + 15\right)$. The square of any real number is always non-negative, meaning $\left(\frac{1}{3}x + 15\right)^2 \geq 0$.

Step 2: Consider the effect of multiplying by $-1$.
When this non-negative square is multiplied by $-1$, the result is always non-positive: $-\left(\frac{1}{3}x + 15\right)^2 \leq 0$.

Step 3: Identify when the function $y$ is less than zero.
For the function to be strictly less than zero, the squared term must be strictly greater than zero: $\left(\frac{1}{3}x + 15\right)^2 > 0$.

Step 4: Determine the zero point to exclude it.
The expression $\left(\frac{1}{3}x + 15\right)^2 = 0$ only when $\frac{1}{3}x + 15 = 0$. Solving this equation gives:

• $\frac{1}{3}x + 15 = 0$
• $\frac{1}{3}x = -15$
• $x = -45$

This means that the function $y = 0$ at $x = -45$. To satisfy $y < 0$, $x$ must be any value other than $-45$.

Therefore, the condition for which the function value is negative is: $x \neq -45$.