At what point on the graph of y=(1/2)x^2-(3/2) is the tangent line parallel to the line 4x-8y=5?
A. (1/2,-3/8)
B. (1/2,-11/8)
C. (2,3/8)
D. (2,1/2)
E. (-1/2,-11/8)
y = (1/2)x^2 - (3/2) -> this is a parabola
differentiate:
dy/dx = (1/2)*2*x - 0
dy/dx = x
The line 4x - 8y = 5 has a slope of,
-8y = -4x + 5
y = (1/2)x - 5/8
slope = 1/2
If this line and the tangent line of the parabola are parallel, they have the same slope:
dy/dx = x
1/2 = x
Substituting this to the parabola equation to get y,
y = (1/2)x^2 - (3/2)
y = (1/2)(1/2)^2 - 3/2
y = 1/8 - 12/8
y = -11/8
Thus point is at (1/2, -11/8).
Hope this helps :3
Well, well, well, looks like we have a math question here! Let me put on my mathematician hat and try to provide you with a clownishly hilarious answer.
To find the point on the graph where the tangent line is parallel to the given line 4x-8y=5, we need to find the derivative of the function y=(1/2)x^2-(3/2). Now, I can explain it in a complicated way using fancy mathematical terms, but I think I'll go with a more fun approach.
Imagine you're trying to climb a hill, and the function y=(1/2)x^2-(3/2) represents the shape of the hill. The tangent line at any point on the hill is like a skateboard ramp, and we want to find the point where that ramp is parallel to the line 4x-8y=5. So, I guess you could say we're looking for the perfect spot to do some sick tricks!
Now, let's get back to business. To find where the tangent line is parallel to 4x-8y=5, we need to find the value of x that satisfies the equation 4x-8y=5. But since we're dealing with parallel lines, we know that their slopes must be equal. So, let's find the slope of the line 4x-8y=5.
Rearranging the equation, we have 8y = 4x - 5, which can also be written as y = (1/2)x - 5/8. And voila, we have our slope! The slope of the line is (1/2).
Now, let's find the derivative of y=(1/2)x^2-(3/2). Taking the derivative, we get dy/dx = x.
Since we want the slopes of the tangent line and the line 4x-8y=5 to be equal, we set the derivative equal to the slope of the line:
x = 1/2
Now, all we have to do is substitute this value of x back into the original equation y=(1/2)x^2-(3/2). Let's do the math!
y = (1/2)(1/2)^2 - (3/2)
y = (1/2)(1/4) - (3/2)
y = 1/8 - 3/2
y = -23/8
So, the point on the graph where the tangent line is parallel to the line 4x-8y=5 is (1/2, -23/8). But wait, none of the answer choices match this result! Oh, the clownery of it all!
To be honest, I'm not sure which answer choice to choose from the given options. The closest one seems to be (1/2, -11/8), but it's not quite right. So, I'll just juggle the options in the air and let you decide, my friend. Good luck!
To find the point on the graph where the tangent line is parallel to the line 4x - 8y = 5, we need to determine the slope of the given line.
First, we'll rewrite the line equation in slope-intercept form (y = mx + b):
4x - 8y = 5
-8y = -4x + 5
y = (1/2)x - 5/8
The slope of this line is 1/2.
Next, we'll find the derivative of the given quadratic function y = (1/2)x^2 - (3/2) to determine the slope of the tangent line at any given point.
Taking the derivative of y with respect to x, we get:
dy/dx = d/dx[(1/2)x^2 - (3/2)]
dy/dx = (1/2)(2x)
dy/dx = x
Now we have the equation for the slope of the tangent line, which is x.
To find the point on the graph where the tangent line is parallel to the line 4x - 8y = 5, we need to equate the slope of the tangent line (x) with the slope of the given line (1/2):
x = 1/2
Solving for x, we get:
x = 1/2
Plugging the value of x back into the quadratic function, we can find the y-coordinate of the point:
y = (1/2)(1/2)^2 - (3/2)
y = 1/8 - 3/2
y = -3/8
Therefore, the point at which the tangent line is parallel to the line 4x - 8y = 5 is (1/2, -3/8).
So, the correct answer is A. (1/2, -3/8).
To find the point on the graph of y=(1/2)x^2-(3/2) where the tangent line is parallel to the line 4x-8y=5, we need to find the derivative of the function y=(1/2)x^2-(3/2).
The derivative of a function represents the slope of the tangent line to the graph of the function at any given point.
Let's find the derivative of y=(1/2)x^2-(3/2) by applying the power rule for derivatives.
The power rule states that the derivative of x^n, where n is a constant exponent, is given by nx^(n-1).
So, using the power rule, the derivative of (1/2)x^2 is:
d/dx [(1/2)x^2] = (1/2)(2)x^(2-1) = x
Therefore, the derivative of y=(1/2)x^2-(3/2) is x.
Now, we need to find the x-coordinate of the point where the tangent line is parallel to the line 4x-8y=5.
The line 4x-8y=5 can be rearranged into the slope-intercept form, y=mx+b, where m is the slope and b is the y-intercept.
Rearranging the equation:
4x-8y=5
-8y=-4x+5
y=(4/8)x-(5/8)
Comparing the slope of this line to the derivative of y=(1/2)x^2-(3/2), we can see that the slope is 4/8, which simplifies to 1/2.
Since parallel lines have the same slope, the x-coordinate of the point on the graph of y=(1/2)x^2-(3/2) where the tangent line is parallel to the line 4x-8y=5 is the solution to the equation x=1/2.
Therefore, the answer is option B: (1/2, -11/8).