graph the parabola y=2x^2-6x-3. plot point on the parabola A[1,-7] and draw a line through A with an angle of inclination equal to 30 degrees. then find the equation of the line and its second point of intersection B, with the parabola
Just take this one step at a time:
a line with inclination 30° has slope tan30° = 1/√3
Now we have a point and a slope, so that line is
(y+7)/(x-1) = 1/√3
y = 1/√3 x - (7+1/√3)
Now find where that line intersects the parabola:
2x^2 - 6x - 3 = 1/√3 x - (7 + 1/√3)
x = 1
x = 2 + 1/(2√3) or 2.288
The second point of intersection is thus
(2 + 1/(2√3) , 1/(2√3) - 41/6)
or (2.288,-6.256)
To graph the parabola y = 2x^2 - 6x - 3, we can plot some points and then connect them to form the curve. We can calculate the y-values for some x-values to find the coordinates of the points on the graph.
Let's calculate the y-value for a few x-values:
For x = -2:
y = 2(-2)^2 - 6(-2) - 3
y = 8 + 12 - 3
y = 17
For x = -1:
y = 2(-1)^2 - 6(-1) - 3
y = 2 + 6 - 3
y = 5
For x = 0:
y = 2(0)^2 - 6(0) - 3
y = 0 - 0 - 3
y = -3
For x = 1:
y = 2(1)^2 - 6(1) - 3
y = 2 - 6 - 3
y = -7
For x = 2:
y = 2(2)^2 - 6(2) - 3
y = 8 - 12 - 3
y = -7
These points can be plotted on a graph with x as the horizontal axis and y as the vertical axis.
On the graph, plot point A (1, -7).
To draw a line through point A with an angle of inclination equal to 30 degrees, we start by finding the slope (m) of that line.
The slope of a line is given by the tangent of the angle of inclination. In this case, since the angle is 30 degrees:
m = tan(30)
m = 1/sqrt(3)
Now we can use the point-slope form of the equation of a line to find the equation of the line passing through point A:
y - y₁ = m(x - x₁)
Replacing the values with those from point A (x₁ = 1, y₁ = -7) and the slope we calculated (m = 1/sqrt(3)):
y - (-7) = 1/sqrt(3)(x - 1)
y + 7 = 1/sqrt(3)x - 1/sqrt(3)
y = 1/sqrt(3)x - 1/sqrt(3) - 7
Simplifying, we can rationalize the denominator:
y = sqrt(3)x/sqrt(3) - sqrt(3)/3 - 7
The equation of the line passing through point A is y = sqrt(3)x/sqrt(3) - sqrt(3)/3 - 7.
To find the second point of intersection B between the line and the parabola, we need to solve the system of equations formed by equating y in the line equation to y in the parabola equation:
sqrt(3)x/sqrt(3) - sqrt(3)/3 - 7 = 2x^2 - 6x - 3
This equation can be simplified to:
x^2 - (5/3)x - 2/3 = 0
Now we can solve this quadratic equation to find the x-coordinates of the intersection points:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 1, b = -(5/3), and c = -(2/3).
x = (-(5/3) ± sqrt((5/3)^2 - 4(1)(-(2/3)))) / (2(1))
After simplifying:
x = (-(5/3) ± sqrt(25/9 + 8/3)) / 2
Using a calculator, we find two possible x-values:
x ≈ -0.31 and x ≈ 2.98
To find the corresponding y-values, we substitute these x-values back into the equation of the parabola:
For x ≈ -0.31:
y ≈ 2(-0.31)^2 - 6(-0.31) - 3
y ≈ 0.191
For x ≈ 2.98:
y ≈ 2(2.98)^2 - 6(2.98) - 3
y ≈ -2.811
Therefore, the second point of intersection B is approximately B(-0.31, 0.191) and B(2.98, -2.811).