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## Main Question or Discussion Point

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

All help would be appreciated!

Thanks,

Kafka

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

y = ax² + bx + c

4 = 0 + 0 + c

c = 4

dy/dx = 2ax + b

-5 = 2ax + b

-5 = 2a(0) + b

-5 = 0 + b

b = -5

*not sure about the bit below*

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 ???

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 ???