Select the graph that represents the function $$ x^{2}-10 x+16 $$

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UpStudy AI · Accepted Answer

1. We start with the function  
   $$
   f(x)=x^{2}-10x+16.
   $$

2. To identify key features, we complete the square. Rewrite $$x^{2}-10x$$ as  
   $$
   x^{2}-10x = \left(x-5\right)^{2}-25.
   $$

3. Substitute back into the function:  
   $$
   f(x)=\left(x-5\right)^{2}-25+16=\left(x-5\right)^{2}-9.
   $$
   This form shows that the vertex of the parabola is at $$(5,-9)$$.

4. Since the coefficient of $$\left(x-5\right)^{2}$$ is positive (namely, $$+1$$), the parabola opens upward.

5. Next, we find the $$x$$-intercepts by setting $$f(x)=0$$:
   $$
   \left(x-5\right)^{2}-9=0.
   $$
   Solve:
   $$
   \left(x-5\right)^{2}=9,
   $$
   $$
   x-5=\pm 3,
   $$
   so
   $$
   x=5\pm3.
   $$
   This gives the intercepts at $$x=2$$ and $$x=8$$.

6. The $$y$$-intercept is found by evaluating $$f(0)$$:
   $$
   f(0)=0^{2}-10\times0+16=16.
   $$
   Thus, the $$y$$-intercept is $$(0,16)$$.

7. The correct graph should show:
   - An upward-opening parabola.
   - A vertex at $$(5,-9)$$.
   - $$x$$-intercepts at $$(2,0)$$ and $$(8,0)$$.
   - A $$y$$-intercept at $$(0,16)$$.

Select the graph that matches these details.
The correct graph is the one that shows an upward-opening parabola with a vertex at (5, -9), $$x$$-intercepts at (2,0) and (8,0), and a $$y$$-intercept at (0,16).

Select the graph that represents the function

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