Determining Positive Values in Quadratic Inequality: y = x² - 6x + 10

Question

Look at the following function:

y=x26x+10 y=x^2-6x+10

Determine for which values of x x the following is true:

f\left(x\right)>0

Step-by-Step Solution

The function y=x26x+10 y = x^2 - 6x + 10 represents a parabola opening upwards since its leading coefficient a=1 a = 1 is positive. Our task is to determine when the function is positive.

First, let's find the vertex of this parabola, which occurs at x=b2a x = -\frac{b}{2a} .

Here, b=6 b = -6 and a=1 a = 1 , 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 f(3)=1 f(3) = 1 , 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, f(x)>0 f(x) > 0 for all xR x \in \mathbb{R} .

Therefore, the function is positive for all values of x x .

Answer

The function is positive for all values of x x .