In a quadratic equation like:
[tex]f(x)=ax^2+bx+c[/tex]To find the vertex use the next formula:
x-coordinate of the vertex:
[tex]h=-\frac{b}{2a}[/tex]y-coordinate of the vertex evaluate the equation for x=h
[tex]f(h)=ah^2+bh+c[/tex]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:
[tex]f(x)=x^2[/tex]Vertex:
[tex]\begin{gathered} a=1,b=0 \\ \\ h=-\frac{0}{2(1)}=0 \\ \\ f(0)=0^2=0 \\ \\ \text{Coordiantes of the vertex:} \\ (0,0) \end{gathered}[/tex]y intercept is 0 (coordinates (0,0))
Zeros: in (0,0)
[tex]\begin{gathered} f(x)=0 \\ x^2=0 \\ x=\sqrt[]{0} \\ x=0 \end{gathered}[/tex]-------------------------------------
[tex]f(x)=x^2+5[/tex]Vertex:
[tex]\begin{gathered} a=1,b=0 \\ \\ h=-\frac{0}{2(1)}=0 \\ \\ f(0)=0^2+5=5 \\ \\ \text{Coordinates of the vertex:} \\ (0,5) \end{gathered}[/tex]y-intercpet is 5 (coordinates (0,5)
Zeros:
[tex]\begin{gathered} f(x)=0 \\ x^2+5=0 \\ x^2=-5 \\ x=\sqrt[]{-5} \\ \end{gathered}[/tex]As the square root of a negative number is not a real number the function has not zeros (doesn't cross the x-axis)
