Given: The value of a and the vertex of the following parabola are given
[tex]\begin{gathered} y=ax²+bx+c \\ a=-2 \\ Vertex=(5,8) \end{gathered}[/tex]Required: To find the x-intercepts of the given parabola.
Explanation: The vertex of the parabola is at (5,8). The x coordinate of the vertex is given by
[tex]x=-\frac{b}{2a}[/tex]Hence, putting x=5 and a=-2 we get,
[tex]\begin{gathered} 5=-\frac{b}{(-4)} \\ b=20 \\ \end{gathered}[/tex]Now the y coordinate of the vertex is 8, and the value of c can be found by putting x=5 in the equation of the parabola.
[tex]\begin{gathered} 8=-2(5)^2+20(5)+c \\ c=-42 \end{gathered}[/tex]Hence the equation of the parabola is
[tex]y=-2x^2+20x-42[/tex]The x-intercepts can be determined by putting y=0 as follows
[tex]-2x^2+20x-42=0[/tex]which gives,
[tex]\begin{gathered} x^2-10x+21=0 \\ (x-7)(x-3)=0 \\ x=7\text{ and} \\ x=3 \end{gathered}[/tex]Following is the graph
Final Answer: The parabola has 2 x-intercepts and the vertex lies above the x-axis.
