Question

please answer to this it is very important.....
please please please please please answer
I will mark as brainlist.......​

Answer 1

AnSwer :-

(a) y = x²- 10x + 24

(i) the cofficient of x² is 1, .i.e., +ve. so the parabola is open upwards ∪

(ii) y = x² - 10x + 24 = 0

(iii) (x - 4)(x - 6) = 0 ; x₁ = 4, x₂ = 6

the parabola cuts the x-axis at x = 4,6

(iv) Vertex :-

dy/dx = 2x - 10 = 0

x = x$_0$ = 5

y = x² - 10x + 24 = 23 - 50 + 24 = -1 = y$_0$

Vertex, P = (5, -1)

|| refer fig-1 from the attachment ||

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(b) y = ax - bx²

(i) the cofficient of x² is -ve, .i.e., the parabola is open downwards ∩

(ii) y = 0 = ax - bx²

x(a - bx) = 0

x₁ = 0,x² = a/b

the parabola cuts the x-axis at x = 0, a/b

(iii) Vertex :-

dy/dx = a - 2bx = 0

x = x$_0$ = a/2b

y = ax - bx² = a²/2b - a²/4b = a²/4b = y$_0$

Vertex :- p = (a/2b, a²/4b)

|| refer fig-2 from the attachment ||

━━━━━━━━━━━━━━━

(c) y = x² + 4

(i) the cofficient of x² is +ve, .i.e., the parabola is open upwards ∪

(ii) y = x² + 4 = 0

x = ±√-4; x is imaginary;

the parabola will not cut the x-axis.

(iii) Vertex :-

dy/dx = 2x = 0

x = x$_0$ = 0

y = x² + 4 = 0 = 0 + 4 = 4 = y$_0$

vertex, P = (0,4)

|| refer fig-3 from the attachment ||

━━━━━━━━━━━━━━━

(d) y = x² - 9

(i) the cofficient of x² is +ve, .i.e., the parabola is open upwards ∪

(ii) y = x² - 9 = 0

x₁ = 3, x₂ = -3

the parabola will cut the x-axis at x = 3,-3

(iii) Vertex :-

dy/dx = 2x = 0

x = 0 = x$_0$

y = x² - 9 = -9 = y$_0$

Vertex, P = (0, -9)

━━━━━━━━━━━━━━━

(e) y = x², y = 3x², y = x²/4

(i) the cofficient of x² is +ve, .i.e., the parabola is open upwards ∪.

(ii) y = x² = 0

x = 0 the parabola will pass through x = 0

(iii) Vertex :-

dy/dx = 2x = 0

x = 0 = x$_0$

y = -x² = 0 = y$_0$

Vertex, P = (0,0)

|| refer fig-5 from the attachment ||

the lower the value of the cofficient of x², the wider the parabola.

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