Given the following Quadratic equation:
[tex]y=-x^2+10x-16[/tex]
1. You can find the vertex as following:
-Find the x-coordinate of the vertex with this expression:
[tex]\frac{-b}{2a}[/tex]
In this case:
[tex]\begin{gathered} a=-1 \\ b=10 \end{gathered}[/tex]
Then, substituting values, you get:
[tex]\frac{-10}{2(-1)}=5[/tex]
- Substitute this value into the Quadratic equation and evaluate, in order to find the y-coordinate of the vertex. This is:
[tex]\begin{gathered} y=-(5)^2+10(5)-16 \\ y=-25+50-16 \\ y=9 \end{gathered}[/tex]
Then, the vertex is:
[tex](5,9)[/tex]
2. Now let's find the roots:
- Substitute the following value of "y" into the equation:
[tex]y=0[/tex]
Then:
[tex]0=-x^2+10x-16[/tex]
- In order to make the leading coefficient positive, multiply both sides of the equation by -1:
[tex]\begin{gathered} (-1)(0)=(-x^2+10x-16)(-1) \\ 0=x^2-10x+16 \end{gathered}[/tex]
- Factor the equation. Find two numbers whose sum is -10 and whose product is 16. These would be -8 and -2. Then:
[tex](x-8)(x-2)=0[/tex]
- The roots are:
[tex]\begin{gathered} x_1=8 \\ x_2=2 \end{gathered}[/tex]
3. Now let's find five points to plot them in the coordinate plane. Give five different values to "x" and evaluate in order to find the corresponding value of "y". Then:
- When
[tex]x=1[/tex]
You get:
[tex]y=-(1)^2+10(1)-16=-7[/tex]
The point is:
[tex](1,-7)[/tex]
- When
[tex]x=3[/tex]
You get:
[tex]y=-(3)^2+10(3)-16=5[/tex]
The point is:
[tex](3,5)[/tex]
- When:
[tex]x=4[/tex]
You get:
[tex]y=-(4)^2+10(4)-16=8[/tex]
The point is:
[tex](4,8)[/tex]
- When:
[tex]x=6[/tex][tex]y=-(6)^2+10(6)-16=8[/tex]
The point is:
[tex](6,8)[/tex]
- When:
[tex]x=7[/tex]
You get:
[tex]y=-(7)^2+10(7)-16=5[/tex]
The point is:
[tex](7,5)[/tex]
Finally, you get the following graph of the equation:
