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

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(x)>0 f\left(x\right)>0

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Step-by-step written solution

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1

Understand the problem

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(x)>0 f\left(x\right)>0

2

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 .

3

Final Answer

The function is positive for all values of x x .

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 \( f\left(x\right) > 0 \).

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