The quadratic function given:
[tex]f(x)=2(x-3)(x+7)[/tex]The x-intercept(s) is the x-axis cutting points. It occurs at y = 0. Thus, we substitute '0' into 'f(x)' and solve for the x values.
[tex]\begin{gathered} f(x)=2(x-3)(x+7) \\ 0=2(x-3)(x+7) \\ x-3=0,x=3 \\ \text{and} \\ x+7=0,x=-7 \end{gathered}[/tex]As we can see, there are 2 x-intercepts, at x = 3 and x = -7.
