Determine the coordinates of the point on the graph of f(x)=3x^2-7x+4 where the tangent line is parallel to the line 5x+y=3
line is
y = -5 x + 3
so slope = -5 = slope of tangent
f' = slope = 6 x - 7
where is that -5?
6 x - 7 = -5
6 x = 2
x = 1/3
so
y = 3(1/9) -7(1/3) + 4
is that 1/9 or 1/3
1/9, since it's 3x^2
don't you read your own equations?
To determine the coordinates of the point on the graph of f(x) where the tangent line is parallel to the given line, we need to find the derivative of f(x) and set it equal to the slope of the given line.
First, let's find the derivative of f(x). The derivative of a quadratic function f(x) = 3x^2 - 7x + 4 can be found by applying the power rule for derivatives. Each term in the function needs to be multiplied by its respective exponent and then subtract 1 from the exponent:
f'(x) = 2 * 3x^(2-1) - 1 * 7x^(1-1) + 0
= 6x - 7
Now we have the derivative of f(x): f'(x) = 6x - 7.
Next, we find the slope of the given line. To do this, we can rewrite the line 5x + y = 3 in slope-intercept form (y = mx + b), where m is the slope of the line:
5x + y = 3
y = -5x + 3
Comparing this form to y = mx + b, we can see that the slope m of the given line is -5.
Since we want the tangent line of f(x) to be parallel to the given line, the slope of the tangent line must also be -5. So we set the derivative equal to -5 and solve for x:
6x - 7 = -5
6x = -5 + 7
6x = 2
x = 2/6
x = 1/3
Now that we have the x-coordinate, we substitute it back into f(x) to find the corresponding y-coordinate:
f(x) = 3x^2 - 7x + 4
f(1/3) = 3(1/3)^2 - 7(1/3) + 4
= 3/9 - 7/3 + 4
= 1/3 - 7/3 + 4
= -6/3 + 4
= -2 + 4
= 2
Therefore, the point on the graph of f(x) = 3x^2 - 7x + 4 where the tangent line is parallel to the line 5x + y = 3 is (1/3, 2).