Question

Write 2 quadratic equations (of any form) that are not equivalent, each with a vertex of (4,5)

Answer 1

`g(x)=-x^2+8x-11\\\\f(x)=x^2-8x+21`

Explanation:

Generally when you're given the vertex of a quadratic, you can express it in vertex form:
`f(x)=a(x-h)^2+k`, where (h, k) is the vertex, and "a" is some constant value that determines the stretch/compression and the direction (up or down)

So we know that: (h, k) = (4, 5), meaning we can plug those values into the equation:
`f(x)=a(x-4)^2+5`

we can change the function by changing the value of the constant "a".

we can use two arbitrary values, but in this example I'll use -1 and 1.

this gives us the following functions:
`f(x)=(x-4)^2+5\\g(x)=-(x-4)^2+5`

these two functions are actually very similar, the only different is they're reflected across the x-axis (and some vertical shift) since the y-value is being negated, or at least part of it.

In your question it states (of any form), so I'm assuming we can leave it in vertex form, but just in case I'll expand them further into standard form

`f(x)=(x-4)^2+5\\\\f(x)=(x-4)(x-4)+5\\\\f(x)=x(x-4)-4(x-4)+5\\\\f(x)=x^2-4x-4x+16+5\\\\f(x)=x^2-8x+21`

We can also expand the other function similarly.

`g(x)=-(x-4)^2+5\\\\g(x)=-[(x-4)(x-4)]+5\\\\g(x)=-[x(x-4)-4(x-4)]+5\\\\g(x)=-[x^2-4x-4x+16]+5\\\\g(x)=-[x^2-8x+16]+5\\\\g(x)=-x^2+8x-16+5\\\\g(x)=-x^2+8x-11`