Chapter 1, Problem 110CR

To determine

To find: Whether the point (-1, 28) lie on the graph of the function $f(x)=x^2$ when it is transformed to $f(x)=-{(x-6)}^2+3$

Answer to Problem 110CR

False, the point (-1, 28) will not lie on the transformation.

Explanation of Solution

Given data:

  $f(x)=x^2$

Calculation:

The parent function $f(x)=x^2$ is moved six units to right and three units upward so the function become $f(x-6)+3={(x-6)}^2+3$ , also the function is reflected in x − axis so $-f(x-6)+3=-{(x-6)}^2+3$

And for checking that (-1, 28) lie on the transformation graph $-f(x-6)+3$

Putting x = -1 should give 28 as result.

So, let $g(x)=-f(x-6)+3$

  $\begin{array}{l}$ g(-1)=-{(-1-6)}^2+3$g(-1)=-49+3\\ g(-1)=-46\ne 28\end{array}$

Thus, the point (-1, 28) will not lie on the transformation.

Conclusion:

(-1,28) is not a point on the graph of $\text{- }f(x-6)+3$