Answer:
The 5 points: (-1, 7), (0, 4), (-2, 4), (-2.5275, 0), (0.5275, 0)
See the graph below
Explanation
Given:
[tex]y\text{ = -3x}^2\text{ - 6x + 4}[/tex]
To find:
plot 5 points on the parabola: the vertex, two points before it, and two points after
First, we need to find the vertex point:
[tex]\begin{gathered} vertex\text{ = \lparen h, k\rparen} \\ h\text{ = }\frac{-b}{2a} \\ k\text{ = f\lparen-b/2a\rparen} \\ \\ a\text{ = -3, b = -6, c = 4} \\ h\text{ = }\frac{-(-6)}{2(-3)}=\text{ 6/-6} \\ h\text{ = -1} \\ \\ k\text{ = f\lparen-b/2a\rparen = f\lparen h\rparen} \\ we\text{ will substitute the value of h into the function} \\ k\text{ = -3\lparen-1\rparen}^2\text{ - 6\lparen-1\rparen + 4} \\ k\text{ = -3\lparen1\rparen + 6 + 4} \\ k\text{ = 7} \\ \\ The\text{ vertex \lparen h, k\rparen = \lparen-1, 7\rparen} \end{gathered}[/tex]
Next, let's find the y-intercept:
it is the value of y when x = 0
[tex]\begin{gathered} y\text{ = -3\lparen0\rparen}^2\text{ - 6\lparen0\rparen + 4} \\ y\text{ = 4} \\ The\text{ point will be \lparen0, 4\rparen} \\ \\ when\text{ x = -2} \\ y\text{ = -3\lparen-2\rparen}^2\text{ -6\lparen-2\rparen + 4} \\ y\text{ = -12 + 12 +4} \\ y\text{ = 4} \\ The\text{ point will be \lparen-2, 4\rparen} \end{gathered}[/tex]
To get the remaining two points, we will use x-intercept
It is the value of x when y = 0
[tex]\begin{gathered} 0\text{ = -3x}^2\text{ - 6x + 4} \\ $x\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$ \\ a\text{ =-3, b = -6, c = 4} \\ \\ x\text{ = }\frac{-(-6)\pm\sqrt{(-6)^2-4(-3)(4)}}{2(-3)} \\ \\ x\text{ = }\frac{6\pm\sqrt{36+48}}{-6}\text{ = }\frac{6\pm\sqrt{84}}{-6} \\ \\ x\text{ = }\frac{6\pm9.165}{-6} \\ \\ x\text{ = }\frac{6+9.165}{-6}\text{ = }\frac{6-9.165}{-6} \\ x\text{ = }\frac{15.165}{-6}\text{ or }\frac{-3.165}{-6} \\ \\ x\text{ = -2.5275 or 0.5275} \end{gathered}[/tex]
The two points: (-2.5275, 0) and (0.5275, 0)
Plotting the points:
