Finding the Equation of a Quadratic Graph

[Norukine]

Hi, I'm currently IBDP AA maths SL, and I was wondering if anyone could give me a hadn't with solving this question. Thank you!

 

[Deleted member 4993]

View attachment 28801
Hi, I'm currently IBDP AA maths SL, and I was wondering if anyone could give me a hadn't with solving this question. Thank you!

When the graph touches x axis at x = x₀, we would get the roots of the equation to be equal.

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.

 

[Otis]

Hint for part (b):

Given: ax^2 + bx + c

x-coordinate of vertex: -b/(2a)

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[Otis]

For part (b), substituting given coordinates (-4,1) and (1,11) into ax^2+bx+c=y yields

16a - 4b + c = 1

a + b + c = 11

The hint in post #3 yields

-8a = b

That's a system of three equations. Continue...

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[HallsofIvy]

View attachment 28801
Hi, I'm currently IBDP AA maths SL, and I was wondering if anyone could give me a hadn't with solving this question. Thank you!

Do you understand what these words mean? For example, (a) says that the graph 'touches the x- axis at 4". First that tells us that when x= 4, y= 0. But the graph "touches" the x-axis, it does NOT cross it! That means that the vertex is at (4, 0). It means that we can write the equation as y= a(x- 4)^2. Since the graph "passes through (2, 12) we have 12= a(2- 4)^2= a(-2)^2= 4a so a= 3. The equation is y= 3(x- 4).^2 We can also write that as y= 3(x^2- 8x+ 16)= 2x^2- 24X+ 48.

(b) is exactly the same. Since the vertex of the parabola is at (-4, 1) we can write it as y= a(x- (-4))^2+ 1= a(x+ 4)^2+ 1. It passes through (1, 4) means that when x= 1, y= 11 so 11= a(1+ 4)^2+ 1. So what is a? And then what is the equation?