Question

Please helpFind the vertex, roots/zeros and y-intercept of the following.

Answer 1

In a quadratic equation like:

`f(x)=ax^2+bx+c`

To find the vertex use the next formula:

x-coordinate of the vertex:

`h=-(b)/(2a)`

y-coordinate of the vertex evaluate the equation for x=h

`f(h)=ah^2+bh+c`

To find the zeros or x-intercepts. Equal the function to zero and solve x.

To find the Y-intercept evaluate the equation for x=0

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For the given equations:

`f(x)=x^2`

Vertex:

`\begin{gathered} a=1,b=0 \\ \\ h=-(0)/(2(1))=0 \\ \\ f(0)=0^2=0 \\ \\ \text{Coordiantes of the vertex:} \\ (0,0) \end{gathered}`

y intercept is 0 (coordinates (0,0))

Zeros: in (0,0)

`\begin{gathered} f(x)=0 \\ x^2=0 \\ x=\sqrt[]{0} \\ x=0 \end{gathered}`

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`f(x)=x^2+5`

Vertex:

`\begin{gathered} a=1,b=0 \\ \\ h=-(0)/(2(1))=0 \\ \\ f(0)=0^2+5=5 \\ \\ \text{Coordinates of the vertex:} \\ (0,5) \end{gathered}`

y-intercpet is 5 (coordinates (0,5)

Zeros:

`\begin{gathered} f(x)=0 \\ x^2+5=0 \\ x^2=-5 \\ x=\sqrt[]{-5} \\ \end{gathered}`

As the square root of a negative number is not a real number the function has not zeros (doesn't cross the x-axis)