Sure! Let's graph the equation [tex]\( y = -x^2 + 12x - 35 \)[/tex].
1. Vertex of the Parabola:
Parabolas have a highest or lowest point called the vertex. For a quadratic function in the form [tex]\( y = ax^2 + bx + c \)[/tex], the x-coordinate of the vertex is given by [tex]\( x = -\frac{b}{2a} \)[/tex].
For our equation [tex]\( y = -x^2 + 12x - 35 \)[/tex]:
- [tex]\( a = -1 \)[/tex]
- [tex]\( b = 12 \)[/tex]
- So, the x-coordinate [tex]\( x = -\frac{12}{2(-1)} = 6 \)[/tex].
To find the y-coordinate of the vertex, plug [tex]\( x = 6 \)[/tex] back into the equation:
[tex]\[
y = -(6)^2 + 12(6) - 35 = -36 + 72 - 35 = 1
\][/tex]
Thus, the vertex is at [tex]\( (6, 1) \)[/tex].
2. Roots (x-intercepts):
To find the x-intercepts, solve for [tex]\( x \)[/tex] when [tex]\( y = 0 \)[/tex]:
[tex]\[
-x^2 + 12x - 35 = 0
\][/tex]
The solutions to this quadratic equation are:
[tex]\[
x = 5 \quad \text{and} \quad x = 7
\][/tex]
So, the roots are [tex]\( (5, 0) \)[/tex] and [tex]\( (7, 0) \)[/tex].
3. Two Additional Points:
We can choose any other two x-values to find additional points. Here, we'll take [tex]\( x = 4 \)[/tex] and [tex]\( x = 8 \)[/tex].
For [tex]\( x = 4 \)[/tex]:
[tex]\[
y = -(4)^2 + 12(4) - 35 = -16 + 48 - 35 = -3
\][/tex]
So, the point is [tex]\( (4, -3) \)[/tex].
For [tex]\( x = 8 \)[/tex]:
[tex]\[
y = -(8)^2 + 12(8) - 35 = -64 + 96 - 35 = -3
\][/tex]
So, the point is [tex]\( (8, -3) \)[/tex].
4. Summary of Points:
- Vertex: [tex]\( (6, 1) \)[/tex]
- Roots: [tex]\( (5, 0) \)[/tex] and [tex]\( (7, 0) \)[/tex]
- Additional Points: [tex]\( (4, -3) \)[/tex] and [tex]\( (8, -3) \)[/tex]
You can plot these points on a coordinate axis:
- [tex]\( (4, -3) \)[/tex]
- [tex]\( (5, 0) \)[/tex]
- [tex]\( (6, 1) \)[/tex]
- [tex]\( (7, 0) \)[/tex]
- [tex]\( (8, -3) \)[/tex]
Once you plot these points, draw a smooth curve passing through them to complete the graph of the equation [tex]\( y = -x^2 + 12x - 35 \)[/tex]. This curve will be a downward-opening parabola.
