Solve y=2x²+16: Finding Values Where Function is Negative

To solve this problem, consider the function $y = 2x^2 + 16$. Our objective is to find values of $x$ for which $y < 0$.

The function, $y = 2x^2 + 16$, is a quadratic function in standard form. Here, $a = 2$, $b = 0$, and $c = 16$.

• Step 1: Determine the Vertex
  Since the quadratic is in the form $y = ax^2 + bx + c$, with $a > 0$, it opens upwards. Therefore, its vertex represents the minimum point. The x-coordinate of the vertex is given by $x = \frac{-b}{2a} = \frac{0}{4} = 0$. Substituting $x = 0$ into the function provides $y = 2(0)^2 + 16 = 16$.
• Step 2: Analyze the Discriminant
  The discriminant $D = b^2 - 4ac = 0^2 - 4 \times 2 \times 16 = -128$. Since $D < 0$, the quadratic has no real roots, meaning it does not cross the x-axis. Instead, it remains entirely above the x-axis.

Since the discriminant is negative, the function has no real roots, confirming that the quadratic function never takes a value below zero. The vertex is at $y = 16$, which is above zero, indicating all function values are positive.

Therefore, the function $y = 2x^2 + 16$ is never less than zero, implying that there are no values of $x$ for which $y < 0$.

Hence, the solution is No x.