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

Aug 28, 2014

The vertex is $\left(0 , 7\right)$.

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:

$f \left(x\right) = - {\left(x - 0\right)}^{2} + 7$

Oct 22, 2014

We can also find the vertex by using the expressions:

$\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$

Standard form:

$a {x}^{2} + b x + c = 0$

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 $\left(0 , 7\right)$

Oct 22, 2014

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

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

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

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

Oct 23, 2014

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.

$f \left(x\right) = 7 - {x}^{2}$

$f ' \left(x\right) = 0 - 2 x$

$f ' \left(x\right) = - 2 x$

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

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

$- 2 x = 0$

$x = 0$

The $x$ value of the vertex is 0.

Now substitute in $x = 0$ in the original function, $f \left(x\right) = 7 - {x}^{2}$

$f \left(0\right) = 7 - {\left(0\right)}^{2}$

$f \left(0\right) = 7$

The vertex is at point $\left(0 , 7\right)$.