the curve y=x^2 - 3x - 4 cross the x-axis at P and Q. the tangent to the curve at P and Q meet at R. the normals to the curve at P and Q meet at S Find the distance RS??

please help i have honestly tried and have had no luck :)

Nn

‘谁会啊 谁来救救我

To find the distance RS, we need to follow these steps:

Step 1: Find the x-coordinates of the points P and Q where the curve intersects the x-axis.
To do this, we set y = 0 and solve for x in the equation y = x^2 - 3x - 4.
0 = x^2 - 3x - 4
Factorizing the equation: (x - 4)(x + 1) = 0
Setting each factor equal to zero:
x - 4 = 0 or x + 1 = 0
x = 4 or x = -1

So the curve crosses the x-axis at P(4, 0) and Q(-1, 0).

Step 2: Find the slopes of the tangent lines at P and Q.
To do this, we take the derivative of the equation y = x^2 - 3x - 4.
dy/dx = 2x - 3
Evaluate the derivative at x = 4 and x = -1:
At P: slope at P (mP) = dy/dx |x=4 = 2(4) - 3 = 8 - 3 = 5
At Q: slope at Q (mQ) = dy/dx |x=-1 = 2(-1) - 3 = -2 - 3 = -5

Step 3: Use the point-slope form to find the equations of the tangent lines at P and Q.
The equation of a line in point-slope form is y - y1 = m(x - x1), where (x1, y1) are the coordinates of the point and m is the slope.
At P: tangent line at P: y - 0 = 5(x - 4) => y = 5x - 20
At Q: tangent line at Q: y - 0 = -5(x - (-1)) => y = -5x - 5

Step 4: Find the intersection point R of the tangent lines.
To find the coordinates of R, we solve the system of equations formed by the two equations of the tangent lines.
Substituting the second equation into the first, we have:
5x - 20 = -5x - 5
10x = 15
x = 3/2
Substituting x = 3/2 into either equation, we get:
y = 5(3/2) - 20 = 15/2 - 20 = -5/2

So the coordinates of R are (3/2, -5/2).

Step 5: Find the slopes of the normal lines at P and Q.
The slope of the normal line is the negative reciprocal of the slope of the tangent line.
At P: slope of normal at P (m'P) = -1/mP = -1/5
At Q: slope of normal at Q (m'Q) = -1/mQ = -1/-5 = 1/5

Step 6: Use the point-slope form to find the equations of the normal lines at P and Q.
Following the same procedure as in Step 3, but using the slopes of the normal lines found in Step 5, we get:
At P: normal line at P: y - 0 = (-1/5)(x - 4) => y = (-1/5)x + 4/5
At Q: normal line at Q: y - 0 = (1/5)(x - (-1)) => y = (1/5)x + 1/5

Step 7: Find the intersection point S of the normal lines.
To find the coordinates of S, we solve the system of equations formed by the two equations of the normal lines.
Substituting the second equation into the first, we have:
(-1/5)x + 4/5 = (1/5)x + 1/5
(-2/5)x = -3/5
x = 3/2
Substituting x = 3/2 into either equation, we get:
y = (-1/5)(3/2) + 4/5 = -3/10 + 4/5 = 19/10

So the coordinates of S are (3/2, 19/10).

Step 8: Find the distance RS.
Using the distance formula √((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) and (x2, y2) are the coordinates of the points R and S, respectively:
Distance RS = √((3/2 - 3/2)^2 + (-5/2 - 19/10)^2)
= √(0^2 + (-55/10)^2)
= √(0 + 3025/100)
= √(3025/100)
= √(3025)/√(100)
= 55/10
= 11/2

Therefore, the distance RS is 11/2.