Respuesta :
We can see the next equation:
[tex]-x^2+4x+5=0[/tex]And this is a parabola with a maximum, since its leading term, in this case, the coefficient of the quadratic term, is negative, and we have to graph the parabola by finding the vertex and plotting it, and by plotting two points at each side of the vertex.
Finding the vertex of the parabola
1. To find the vertex, we can use the next formulas:
[tex]\begin{gathered} \text{ Finding the x-value of the vertex:} \\ \\ x_v=-\frac{b}{2a} \\ \\ \text{ Finding the y-value of the vertex:} \\ \\ y_v=c-\frac{b^2}{4a} \end{gathered}[/tex]And we can apply this to any polynomial of the form:
[tex]\begin{gathered} ax^2+bx+c \\ \end{gathered}[/tex]2. Then the vertex of this parabola is:
The x-value of the vertex is:
[tex]\begin{gathered} \begin{equation*} -x^2+4x+5 \end{equation*} \\ \\ a=-1,b=4,c=5 \\ \\ x_v=-\frac{4}{2(-1)}=-\frac{4}{-(2)}=2 \\ \\ x_v=2 \end{gathered}[/tex]The y-value of the vertex is:
[tex]\begin{gathered} y_v=c-\frac{b^{2}}{4a},\Rightarrow a=-1,b=4,c=5 \\ \\ y_v=5-\frac{4^2}{4(-1)}=5-\frac{16}{-4}=5+4=9 \\ \\ y_v=9 \\ \end{gathered}[/tex]Therefore, the vertex of the parabola is given by (2, 9).
Graphing the parabola
3. Now, to find the solutions to the parabola, we have to find the values of x which result in zero. They are called the zeros of the equation, and they happen when the parabola crosses the x-axis, and when y = 0.
4. Then we need two points at the left of the vertex (at the left of x = 2), and two points at the right of x = 2), and they could be: x = -2, x = 0, x = 4, x = 6.
5. Then we have to obtain the corresponding values for y using the original equation of the parabola:
For x = -2
[tex]\begin{gathered} f(x)=-x^2+4x+5 \\ \\ f(-2)=-(-2)^2+4(-2)+5 \\ \\ f(-2)=-(4)-8+5=-12+5=-7 \\ \\ \text{ Therefore, we have the following coordinate to graph the parabola} \\ \\ (-2,-7) \end{gathered}[/tex]And we can proceed in a similar way with the other points as follows:
For x = 0
[tex]\begin{gathered} f(x)=-x^{2}+4x+5 \\ \\ f(0)=-(0)^2+4(0)+5=5 \\ \\ f(0)=5\Rightarrow(0,5)\rightarrow\text{ This is the other point.} \end{gathered}[/tex]For x = 4
[tex]\begin{gathered} f(x)=-x^{2}+4x+5 \\ \\ f(4)=-(4)^2+4(4)+5=-16+16+5 \\ \\ f(4)=5 \\ \\ \text{ Then the coordinates of the point is \lparen4, 5\rparen} \end{gathered}[/tex]For x = 6
[tex]\begin{gathered} f(x)=-x^{2}+4x+5 \\ \\ f(6)=-(6)^2+4(6)+5=-36+24+5=-7 \\ \\ f(6)=-7 \\ \\ \text{ Then the other point is \lparen6, -7\rparen} \end{gathered}[/tex]6. Now, we can graph the parabola using the coordinates of the vertex, and four additional points, two on each side of the vertex as follows:
• Vertex = (2, 9), and the points (-2, -7), (0, 5), (4, 5), (6, -7)
7. Now, to find the solutions to the equation, we have to find the points where the parabola passes through the x-axis, that is, when y = 0. Then, we can see that these points are:
Therefore, we can see that we have two solutions for the values of x, x = -1, and x = 5, since the result in the original equation will be 0 as follows:
[tex]\begin{gathered} x=-1\Rightarrow \\ \\ f\mleft(-1\mright)=-(-1)^2+4(-1)+5=-1-4+5=0 \\ \\ f(-1)=0 \\ \\ \text{ And also:} \\ \\ x=5 \\ \\ f(5)=-(5)^2+4(5)+5=-25+20+5=-5+5=0 \\ \\ f(5)=\text{ 0} \end{gathered}[/tex]Therefore, in summary, the two solutions obtained from the graph, that are solutions of the equation are x = -1, x = 5, and if we separate them by using commas, we can rewrite them as follows:
[tex]\begin{gathered} \text{ Solutions}=\lbrace x=-1,x=5\rbrace \\ \end{gathered}[/tex]
