Plot the graph of the function $y=-2x^2+8x-5$
To plot the graph of the quadratic function $y=-2x^2+8x-5$, follow these steps:
- Determine the coordinates of the parabola's vertex
The function has the form $y=a(x-h)^2+k$, where $(h,k)$ is the vertex coordinates.
For our function $y=-2x^2+8x-5$, the coefficient $a=-2$.
The $x$ coordinate of the vertex: $h = -\frac{b}{2a} = -\frac{8}{2(-2)} = -\frac{8}{-4} = 2$
The $y$ coordinate of the vertex: $k = f(h) = -2(2)^2+8(2)-5 = -2(4)+16-5 = -8+16-5 = 3$
Thus, the vertex of the parabola is at the point $(2,3)$.
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Determine the direction of the parabola's branches
Since the coefficient $a=-2$ is negative, the branches of the parabola are directed downward.
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Find the points of intersection with the coordinate axes
With the $y$ axis (at $x=0$):
$y = -2(0)^2+8(0)-5 = -5$
The point of intersection with the $y$ axis: $(0,-5)$
With the $x$ axis (at $y=0$):
$0 = -2x^2+8x-5$
$2x^2-8x+5 = 0$
Use the discriminant formula: $D = b^2-4ac = (-8)^2-4(2)(5) = 64-40 = 24$
$x_{1,2} = \frac{8 \pm \sqrt{24}}{4} = \frac{8 \pm 2\sqrt{6}}{4} = 2 \pm \frac{\sqrt{6}}{2}$
$x_1 \approx 3.22$ and $x_2 \approx 0.78$
The points of intersection with the $x$ axis: $(0.78,0)$ and $(3.22,0)$
- Plot the graph
The graph is a parabola with a vertex at the point $(2,3)$, branches directed downward, intersecting the $y$ axis at the point $(0,-5)$, and the $x$ axis at the points $(0.78,0)$ and $(3.22,0)$.
