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!

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