Given:
[tex]f(x)=4x^2-8x-4[/tex]
The x-intercepts are the values of x where the function or y-coordinate is zero.
Therefore, the x - intercepts are given when f(x)=0
Hence putting f(x)=0,we get:
[tex]\begin{gathered} f(x)=0 \\ 4x^2-8x-4=0 \end{gathered}[/tex]
The solutions of quadratic equation are given as:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Here, a = 4 b = -8 c = -4,
[tex]\begin{gathered} x=\frac{8\pm\sqrt[]{64-(4\times-4\times4)}}{2\times4} \\ =\frac{8\pm\sqrt[]{64+64}}{8} \\ =\frac{8\pm\sqrt[]{128}}{8} \\ =\frac{8+\sqrt[]{128}}{8},\frac{8-\sqrt[]{128}}{8} \\ =\frac{8+11.3137}{8},\frac{8-11.3137}{8} \\ =\frac{19.3137}{8},\frac{-3.3137}{8} \\ =2.4142125,-0.4142125 \\ =2.4,-0.4 \end{gathered}[/tex]
Hence, the x intercepts are ( - 0.4,0 ) and ( 2.4,0)
Since the graph is an upward parabola, it will have a minimum value at its vertex that is (1,-8) and hence, the minimum value of f(x) is - 8.
