Answer:
[tex]\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}[/tex]
Step-by-step explanation:
x-intercepts are when the curve intercepts the x-axis, so when y =0.
Therefore, to find the x-intercepts, substitute y = 0 and solve for x.
The vertex is the turning point: the minimum point of a parabola that opens upward, and the maximum point of the parabola that opens downward. As a parabola is symmetrical, the x-coordinate of the vertex is the midpoint of the x-intercepts.
Equation: [tex]y=x(x-2)[/tex]
[tex]\implies x(x-2)=0[/tex]
[tex]\implies x=0[/tex]
[tex]\implies (x-2)=0 \implies x=2[/tex]
Therefore, the x-intercepts are x = 0 and x = 2
The midpoint of the x-intercepts is x = 1, so the x-coordinate of the vertex is x = 1
Equation: [tex]y=(x-4)(x+5)[/tex]
[tex]\implies (x-4)(x+5)=0[/tex]
[tex]\implies (x-4)=0 \implies x=4[/tex]
[tex]\implies (x+5)=0 \implies x=-5[/tex]
Therefore, the x-intercepts are x = -5 and x = 4
The midpoint of the x-intercepts is x = -0.5, so the x-coordinate of the vertex is x = -0.5
Equation: [tex]y=-5x(3-x)[/tex]
[tex]\implies -5x(3-x)=0[/tex]
[tex]\implies -5x=0 \implies x=0[/tex]
[tex]\implies (3-x)=0 \implies x=3[/tex]
Therefore, the x-intercepts are x = 0 and x = 3
The midpoint of the x-intercepts is x = 1.5, so the x-coordinate of the vertex is x = 1.5
[tex]\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}[/tex]
