Question 622·Hard·Math

A translation moves each point without reshaping the graph. To move $p$ right by $1$, use $x-1$ as its input: at the new position $x=1$, the old input is $0$. Moving up $5$ adds $5$ to the output, so:

A right shift changes the input; an up shift changes the output.

Type $p(x)=x^4-8x^2+7$, then $q(x)=p(x-1)+5$. Desmos draws the new graph with two crossings to the right of the $y$-axis. Each crossing is an $x$-intercept, where the output is $0$. The crossings left of that axis have negative $x$-values, so they can't answer the question.

Click the left of the two positive crossings. Desmos labels it approximately $(2.414,0)$. Type $1+\sqrt{2}$ on a new line; Desmos prints approximately $2.414$. So the least positive $x$-intercept of $q$ is $1+\sqrt{2}$. Choice D.

For an exact result, let $u=x-1$, the input to $p$ after the shift. An $x$-intercept has $q(x)=0$. Substitute the given rule for $p(u)$, then combine $7$ and $5$:

The equation contains only even powers of $u$. Let $v=u^2$, so $u^4=v^2$. Replace those powers to get a quadratic equation, one whose highest power is $2$:

The numbers $2$ and $6$ multiply to $12$ and add to $8$. Use them to factor the quadratic:

A product is $0$ when at least one factor is $0$. Set each factor to $0$:

Add the constant in each case:

Put $u^2$ back in place of $v$:

Take both square roots in each case, since a positive and a negative number have the same square:

Since $u=x-1$, add $1$ to each old input to get the new $x$-values:

Both minus-sign values are negative because each square root exceeds $1$. Of the positive values, $1+\sqrt{2}$ is smaller than $1+\sqrt{6}$. So the least positive $x$-intercept is $1+\sqrt{2}$. Choice D.