Verified answer

1) y = ax^2 + bx + c

y' = 2ax + b = 0

x = -b/2a

a(b^2 / 4a^2) - b^2/2a + c = 6

b^2 / 4a - b^2/2a + c = 6

(b^2 - 2b^2)/4a + c = 6

-b^2/4a + c = 6

4a - 2b + c = 0

16a + 4b + c = 0

solving 3 equations 3 unknowns yields:

a = -2/3 , b = 4/3 , c = 16/3

y = 2/3(-x^2 + 2x + 8)

2) Using the vertex formula(can also be solved same as above):

y = a(x - h)^2 + k => equation of parabola-vertex form, where (h , k) is the vertex:

in this case vertex is the minimum hence a > 0 and k = -5, then:

y = a(x - h)^2 - 5 => plug-in the given x-intercepts:

a(2 + h)^2 - 5 = 0

a(3 - h)^2 - 5 = 0

2 + h = 3 - h

2h = 1 => h = 1/2

a(5/2)^2 = 5

25a = 20 => a = 4/5

y = 4/5(x - 1/2)^2 - 5

Edit: The answer below for problem 1 is incorrect:

y = ax^2 + bx + c => equation of parabola in standard form, if a > 0 , parabola opens upward which makes the vertex the minimum value, if a < 0 , it open down, hence the vertex is at maximum.

To find the coordinates of the vertex you can use:

x = -b/2a

y = f(-b/2a)

for your equation:

y = -x^2 + 2x + 8

x = -2/-2 = 1

y(1) = -1 + 2 + 8 = 9 => so your equation has a vertex at(1 , 9), which gives you a maximum at y=9 and not y = 6, therefore the equation is completely incorrect!

Edit: The above answer uses differentiation, an advanced level technique, to identify the value of x at the minimum/maximum point in the curve. Although this is a valid option, it is not something taught at an elementary level so cannot possibly be used in answer to a GCSE level question. By observing the symmetry of the parabolic curve, we can arrive at the same conclusion without depending on the fact that the gradient at that point (the first differential) must be zero.

1.

x intercepts at -2 and 4, so y = 0 at (x = -2 and x = 4)

so parabola is of the form:

y = A * (x + 2) * (x - 4)

To find A:

By symmetry, the maximum / minimum value of y must be at a value of x half way between the zero crossings (the points there it crosses the x axis).

Therefore the maximum value of y must be at x = (-2 + 4) / 2 = 1

at x = 1, y = A * (1 + 2) * (1 - 4)

y = A * 3 * (-3)

y = A * (-9)

A = - y / 9

the maximum value of y is 6, so

A = - 6 / 9

A = -2/3

Substituting back into the original equation we have

y = -2/3 * (x + 2) * (x - 4)

y = - 2/3x^2 + 4/3x + 16/3

2)

y = A * (x + 2) * (x - 3)

at minimum value of y, x = (-2 + 3) / 2 = 1/2

y = A * (1/2 + 2) * (1/2 - 3)

y = A * (5/2) * (-5/2)

the minimum value of y is -5 so

-5 = A * (- 25 / 4)

A = 4 / 5

Substitute back into original equation

y = 4/5 * (x + 2) * (x - 3)

y = 4/5 x^2 - 4/5 * x - 24/5

3)

y = A * (x + 6) * x

max point at x = (-6 + 0) / 2 = -3

y = A * (-3 + 6) * -3

y = A * (-9)

A = y / (-9)

max point at y = 9

A = 9 / (-9)

A = -1

substituting back into original equation

y = -1 * (x + 6) * x

y = -x^2 - 6x

Edit: fixed answer to Q1 (copying errors)