To graph the quadratic function [tex]\( y = -2x^2 + 12x - 14 \)[/tex], follow these steps:
1. Find the vertex of the parabola:
The vertex form of a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex] can be derived using the formulas:
[tex]\[
h = -\frac{b}{2a} \quad \text{and} \quad k = c - \frac{b^2}{4a}
\][/tex]
Here, [tex]\( a = -2 \)[/tex], [tex]\( b = 12 \)[/tex], and [tex]\( c = -14 \)[/tex].
The x-coordinate of the vertex [tex]\( h \)[/tex] is:
[tex]\[
h = -\frac{12}{2(-2)} = 3
\][/tex]
The y-coordinate of the vertex [tex]\( k \)[/tex] is:
[tex]\[
k = -14 - \frac{12^2}{4(-2)} = 4
\][/tex]
Hence, the vertex of the parabola is at the point [tex]\( (3, 4) \)[/tex].
2. Plot the vertex on the graph:
Plot the point [tex]\( (3, 4) \)[/tex] on the Cartesian plane. This point is the highest point on the graph since the parabola opens downward due to the negative coefficient [tex]\( a \)[/tex].
3. Find a second point on the parabola:
To find another point, choose a convenient x-value, for instance, [tex]\( x = 0 \)[/tex]:
[tex]\[
y = -2(0)^2 + 12(0) - 14 = -14
\][/tex]
This gives us the point [tex]\( (0, -14) \)[/tex].
4. Plot the second point on the graph:
Plot the point [tex]\( (0, -14) \)[/tex] on the Cartesian plane.
5. Draw the parabola:
Using the vertex [tex]\( (3, 4) \)[/tex] and the second point [tex]\( (0, -14) \)[/tex], sketch the parabola. Since the parabola is symmetric about the vertical line [tex]\( x = 3 \)[/tex], ensure that the parabola opens downward and passes through these points.
With these steps, you have successfully plotted the quadratic function [tex]\( y = -2x^2 + 12x - 14 \)[/tex] on the graph. The vertex is at [tex]\( (3, 4) \)[/tex], and another point on the parabola is [tex]\( (0, -14) \)[/tex].
