Answer:
(a)(1.5, -3)
(b)(0,6)
(c)(2.37, 0), (0.63, 0)
Explanation:
Given the equation of the parabola:
[tex]f\mleft(x\mright)=4x^2-12x+6[/tex]Part A
To find the vertex, first, determine the equation of the line of symmetry:
[tex]\begin{gathered} x=-\frac{b}{2a} \\ =-\frac{-12}{2\times4}=\frac{12}{8} \\ x=1.5 \end{gathered}[/tex]Substitute x=1.5 to find f(1.5).
[tex]f(1.5)=4(1.5)^2-12(1.5)+6=-3[/tex]The vertex of the parabola: (1.5, -3)
Part B
The vertical intercept is the point where x=0
[tex]\begin{gathered} f\mleft(0\mright)=4(0)^2-12(0)+6 \\ f(0)=6 \end{gathered}[/tex]The vertical intercept is the point (0,6).
Part C
The x-intercepts are the point where f(x)=0.
[tex]f\mleft(x\mright)=4x^2-12x+6=0[/tex]From the equation above: a=4, b=-12, c=6
Using the quadratic formula:
[tex]\begin{gathered} x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a} \\ =\dfrac{-(-12)\pm\sqrt[]{(-12)^2-4(4)(6)}}{2\times4} \\ =\dfrac{12\pm\sqrt[]{144-96}}{8} \\ =\dfrac{12\pm\sqrt[]{48}}{8} \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} x=\dfrac{12+\sqrt[]{48}}{8}\text{ or }x=\dfrac{12-\sqrt[]{48}}{8} \\ x=2.37\text{ or }x=0.63 \end{gathered}[/tex]The coordinates of the two x-intercepts of the parabola are:
• (2.37, 0)
,• (0.63, 0)
