Test 3 Section 3 Number 10 please

There are a couple ways to think about this one. First, you can plug 25 in for y and solve for the two x values:

    \begin{align*}y&=(x-11)^2\\25&=(x-11)^2\\5&=\pm (x-11)\end{align*}

From there, you know that either 5=x-11 or 5=11-x, so x must be 16 or 6. 16 and 6 are 10 apart, so the line that connects the two intersections of the y=(x-11)^2 parabola and the y=25 line must be 10 apart.

t3s3-10

The other way to go is more abstract. Basically, you can think about the general shape of a parabola with a leading coefficient of 1, and the fact that y=(x-11)^2 is really just the y=x^2 parabola shifted 11 units to the right. A right or left shift won’t change the distance between the two points that intersect the y=25 graph at all, so you can think about the y=x^2 parabola instead. You know that’s going to hit y=25 when x equals 5 or –5, and 5 and –5 are 10 units apart.

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