Once you have an expression for $q(x)$, what equation must you solve to find the $x$-intercepts? Remember what $y$ equals at an $x$-intercept.

After simplifying $q(x)$, you will get a 4th-degree polynomial. Try factoring it as a product of two quadratics by matching coefficients, especially the constant term 5.

After solving the two quadratics with the quadratic formula, you will get four roots. Which of these are positive, and which of the positive ones is the smallest?

In Desmos, type p(x) = x^4 - 8x^2 + 7. Then define the translated function using the rule for shifts: type q(x) = p(x - 1) + 5.

Click on the graph of q(x) where it crosses the x-axis; Desmos will show the x-coordinates of the intercepts. Identify all positive x-intercepts and then determine which of those positive values is the smallest.

A translation 1 unit right replaces $x$ with $x-1$ inside the function, and a translation 5 units up adds 5 to the output.

So the new function is

Since $p(x) = x^4 - 8x^2 + 7$, we have

First expand $(x-1)^2$ and $(x-1)^4$:

• $(x-1)^2 = x^2 - 2x + 1$
• $(x-1)^4 = (x^2 - 2x + 1)^2 = x^4 - 4x^3 + 6x^2 - 4x + 1$

Now substitute into $q(x)$:

To find the $x$-intercepts, set $q(x)=0$:

Try factoring the quartic as a product of two quadratics:

Matching the constant term $bd = 5$ and the $x^4$ term suggests trying integer pairs whose product is 5. With some trial, we find

So the equation becomes

Set each factor equal to zero and solve:

1. For $x^2 - 2x - 1 = 0$:

2. For $x^2 - 2x - 5 = 0$:

Now simplify the square roots:

• $\sqrt{8} = 2\sqrt{2}$, so the first quadratic gives $x = 1 \pm \sqrt{2}$.
• $\sqrt{24} = 2\sqrt{6}$, so the second quadratic gives $x = 1 \pm \sqrt{6}$.

The positive $x$-intercepts are $1+\sqrt{2}$ and $1+\sqrt{6}$; among these, the least positive value is $1+\sqrt{2}$, which matches answer choice D.