How do you find the intercepts, vertex and graph #f(x)=x^2+4#?

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Answer 1 · 2017-12-31T06:26:45

Vertex: #(0,4)#
x-intercept: doesn't exist
y-intercept: #(0,4)#

The x-vertex of a quadratic equation #y=ax^2+bx+c# is given by #-b/(2a)#.

In the equation #y=x^2+4#, #a=1#, #b=0#, and #c=4#

Therefore, the x-vertex is #-0/2=0#.

Plugging in #x=0# in to the equation gives us the y-vertex.

#f(0)=0^2+4=4#

Therefore, the y-vertex is #4#.

Combine the x-vertex and the y-vertex to get the vertex coordinates.

Vertex: #(0,4)#

The x-intercept is given by plugging in #y=0# into the original equation.

#0=x^2+4#

#x^2=-4#

#x=sqrt(-4)=2i#, #xnotin RR#

Therefore, there is no x-intercept.

The y-intercept is given by plugging in #x=0# into the equation.

#y=0^2+4=4#

Therefore, the y-intercept is at #(0,4)#.

Here is the graph for #f(x)=x^2+4#:

graph{x^2+4 [-20.17, 19.83, -1.2, 18.8]}

I hope that helps!