Given: The value of 'a' in y = ax²+bx+c is a=2 and vertex is at (2,0).
Required: To find the x-intercepts.
Explanation: The x-coordinate of the vertex is 2. Also, we know that x-coordinate is given by
[tex]x=-\frac{b}{2a}[/tex]Hence, putting the value of x=2 and a=1 we get
[tex]\begin{gathered} 2=-\frac{b}{2(1)} \\ b=-4 \end{gathered}[/tex]Now putting y=0, x=2, a=1, and b=-4 in eq of parabola we get
[tex]\begin{gathered} 0=2^2-4(2)+c \\ c=4 \end{gathered}[/tex]Now the equation of the parabola is,
[tex]y=x^2-4x+4[/tex]Now to find x-intercepts put y=0 i.e.,
[tex]\begin{gathered} x^2-4x+4=0 \\ (x-2)^2=0 \\ x=2,2 \end{gathered}[/tex]Hence there is only one x-intercept at (2,0). The opening of the parabola can be seen in the graph below-
Final Answer: The parabola has one x-intercept because the parabola opens upward and the vertex is on the x-axis.