Look at the following function:
Determine for which values of the following is true:
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Look at the following function:
Determine for which values of the following is true:
The function represents a parabola opening upwards since its leading coefficient is positive. Our task is to determine when the function is positive.
First, let's find the vertex of this parabola, which occurs at .
Here, and , so:
\begin{align*} x_{vertex} &= -\frac{-6}{2 \times 1} \\ &= \frac{6}{2} \\ &= 3. \end{align*}Next, we evaluate the function at this vertex:
\begin{align*} f(3) &= 3^2 - 6 \cdot 3 + 10 \\ &= 9 - 18 + 10 \\ &= 1. \end{align*}Since , which is greater than zero, we observe that at the vertex the function is indeed positive.
Moreover, because the parabola opens upwards and the vertex value is positive, the entire parabola lies above the x-axis. Consequently, for all .
Therefore, the function is positive for all values of .
The function is positive for all values of .
The graph of the function below does not intersect the \( x \)-axis.
The parabola's vertex is marked A.
Find all values of \( x \) where
\( f\left(x\right) > 0 \).
The discriminant is negative. This means the parabola never touches the x-axis - it stays completely above it!
Check two things: 1) Does it open upward? (positive leading coefficient) 2) Is the vertex value positive? If both are yes, the function is always positive!
If (opens downward), you'd need the vertex value to be negative for the function to be always negative. With a positive vertex, a downward parabola would cross the x-axis twice.
Testing values helps, but it's not reliable! You might miss the minimum point. Always find the vertex systematically using for the complete picture.
For upward parabolas, the vertex gives the minimum value of the function. If this minimum is positive, then every other point on the parabola must also be positive!
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