How to Determine the Equation of a Parabola with Given Points and Gradient?

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SUMMARY

The equation of the curve defined by the points and gradient provided is y = x² - 5x + 4. The constants were determined as follows: c = 4 from the y-intercept (0,4), b = -5 from the gradient at that point, and a = 1 using the x-intercept (1,0). This process involved substituting known values into the standard quadratic form and solving for the unknowns.

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Kafka
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Hi there,

I have to complete the following question, but I have no idea how to approach it (there are four other parts to the question that I can't complete until I get the first bit). I have made numerous attempts, but am not sure how to get the a part of the equation.

All help would be appreciated!

Thanks,
Kafka

Question
A curve f(x) is defined by the equation : y = ax² + bx + c, where a, b and c are constants.

The curve crosses the y-axis at the point (0,4). At this point the gradient of the graph is -5.

The curve crosses the x-axis at point (1,0).

(i) Find the values of a, b, and c and write down the equation of the curve

Attempt

Sub point (0.4) into equation to get c (x=0,y=4):

y = ax² + bx + c
4 = 0 + 0 + c
c = 4

If gradient at point (0,4) is -5, then dy/dx must be equal to -5.

dy/dx = 2ax + b
-5 = 2ax + b
-5 = 2a(0) + b
-5 = 0 + b
b = -5

*not sure about the bit below*

c=-5, b=4, so sub these into equation of curve and use a point to find a

y = ax² + bx + c

at (1,0) x=1, y=0

y = ax² + bx + c
0 = a1² + (-5 x 1) + 4
0 = a -5 + 4
0 = a - 1
a = 1 ?
 
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Looks good to me :approve:
 
Now if only PHYSICS students will recognize this as a projectile problem...
 
Kafka, you did that exactly right.

It does, however, have nothing to do with "differential equations" so I am moving it to the Calculus section.
 
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

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