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

4 Answers

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#

Oct 22, 2014

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.

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)#.

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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(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.