Question

What is the vertex of `f(x)=7-x^2`?

Answer 1

The vertex is `(0,7)`.

Sometimes we have problems with the easier questions because it's not exactly in the form we're used to. Normally for a quadratic, you would complete the square to find the vertex. But this quadratic is already in vertex or standard form:

`f(x)=-(x-0)^2+7`

Answer 2

We can also find the vertex by using the expressions:

`(-b/(2a), f(-b/(2a)))`

Standard form:

`ax^2+bx+c=0`

In this example, `a=−1` and `b=0`

`x=−0/(2(−1))=0/2=0`

`y=f(0)=7−0^2=7`

Same result of `(0,7)`

See a video example of various methods.

Answer 3

Yes, the vertex is at (0,7), but I would like to address this problem graphically.

The graph of a function `f(x)=x^2` is a parabola with branches directed upward and a vertex at the point `(0,0)`, as everybody knows.

When you consider a graph of a function `f(x)=-x^2`, you just turn the graph of `f(x)=x^2` upside down. The vertex will still be at `(0,0)`, but the branches of a parabola will be directed downwards.

Next we transform our function into `f(x)=7-x^2`, which adds `7` to all values of a function `f(x)=-x^2`. That shifts an entire graph by `7` upwards. Vertex also gets shifted by `7`, so its position is shifted from `(0,0)` to `(0,7)`.

Answer 4

We could also use Calculus to solve this question.

We have to recognize that this is a quadratic equation which is just a parabola.

We know that a parabola will have either a maximum or minimum at the vertex.

The derivative of a function is the slope of the tangent line at a specific point on the function.

The derivative or tangent line at the vertex will have a slope of 0.

`f(x)=7-x^2`

`f'(x)=0-2x`

`f'(x)=-2x`

`-2x` is the derivative, the slope of the tangent line.

Set the derivative equal to zero to find the `x` value at the vertex.

`-2x=0`

`x=0`

The `x` value of the vertex is 0.

Now substitute in `x=0` in the original function, `f(x)=7-x^2`

`f(0)=7-(0)^2`

`f(0)=7`

The vertex is at point `(0,7)`.

See a video example of various methods.