Question

With explanation state a possible vertex form quadratic function with the stated characteristics.

Answer 1

Step 1

A)The given function is congruent to;

`y=x^2`

with the graph

For the vertex to open downwards. it means the required function will be a reflection of y=x².

There in the transformation to get the first step will be;

`f(x)=-x^2`

If the vertex is on the left of the y-axis then we will have the vertex form to be;

`\begin{gathered} y=a(x-h)^2+k \\ The\text{ given function has vertex\lparen0,0\rparen} \\ With\text{ x=2, y=4} \\ 4=a(2-0)^2+0 \\ 4=4a \\ (4a)/(4)=(4)/(4) \\ a=1 \\ Hence\text{ the vertex form of y=x}^2\text{ is y=1\lparen x-0\rparen}^2+0 \end{gathered}`

The vertex form of the part A will be;

Answer to the part A

`y=-1(x+2)^2+2`

Step 2

If the function opens upwards, wider than y=x², with vertex below the x-axis and to the right of the y-axis, the function will be;

Answer to the part B;

`y=(1)/(2)(x-2)^2-2`