Question

For the function : f(x) = -(x-2)^2+9 name all transformations, intercepts and the vertex

Answer 1

Given:

The given function is:

$f(x)=-(x-2)^2+9$

To find:

The transformations, intercepts and the vertex.

Solution:

The vertex form of a parabola is:

$y=a(x-h)^2+k$         ...(i)

Where, a is a constant and (h,k) is vertex.

If a<0, then the graph of parent quadratic function $y=x^2$ reflect across the x-axis.

If h<0, then the graph of parent function shifts h units left and if h>0, then the graph of parent function shifts h units right.

If k<0, then the graph of parent function shifts k units down and if k>0, then the graph of parent function shifts k units up.

We have,

$f(x)=-(x-2)^2+9$          ...(ii)

On comparing (i) and (ii), we get

$a=-1,h=2,k=9$

So, the graph of the parent function reflected across the x-axis, and shifts 2 units right and 9 units up.

Putting x=0 in (ii), we get

$f(0)=-(0-2)^2+9$

$f(0)=-4+9$

$f(0)=5$

The y-intercept is 5.

Putting f(x)=0 in (ii), we get

$0=-(x-2)^2+9$

$(x-2)^2=9$

Taking square root on both sides, we get

$(x-2)=\pm \sqrt{9}$

$x=\pm 3+2$

$x=3+2\text{ and }x=-3+2$

$x=5\text{ and }x=-1$

Therefore, the x-intercepts are -1 and 5.

The values of h and k are 2 and 9 respectively and (h,k) is the vertex of the parabola.

Therefore, the vertex of the parabola is (2,9).