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

User Abk
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1 Answer

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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 √(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).

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