Question

please answer fast ......​

Answer 1

Answer :

  $y = -3x^2-4x+3$

Explanation :

The standard form of a parabola is

  $y = ax^2 + bx + c,$

where $a \ne 0$, and $a,b,c$ are real numbers.

If it passes through (0,3) then when x = 0, y = 3 so this means that

  $3 = a(0)^2 + b(0) + c \implies c = 3$

so $y = ax^2 + bx + 3$.

If it passes through (1,-4), then when x = 1, y = -4 so

  $\begin{aligned}-4 &= a(1)^2 + b(1) + 3 \\a+b+3 &= -4 \\a+b &= -7 && \text{(I).}\end{aligned}$

If it passes through (-1,4) then when x = -1, y = 4 so

  $\begin{aligned}4 &= a(-1)^2 + b(-1) + 3 \\a-b+3 &= 4 \\a-b &= 1 && \text{(II).}\end{aligned}$

Because both (I) and (II) need to be satisfied, we have the system of equations,

  $\begin{cases}a+b &= -7\qquad\text{(I)}\\a-b &= 1\qquad\text{(II)}\end{cases}$

which we can easily solve by adding the two equations up to get

  $\begin{aligned}(a+a) + (b-b) &= -7 + 1 \\ 2a&= -6 \\a &= -3.\end{aligned}$

Then we take any of the previous equations to solve for b:

  $\begin{aligned}a+b &= -7\\-3 + b &= -7 \\ b &= -4\end{aligned}$

Thus the parabola in standard form is

  $y = -3x^2-4x+3.$

Answer 2

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