Question

x | y\
-3 | 0\
-2 | 6\
-1 | 2\
0 | 6\
1 | 0\
\
formula x = ______

Explanation:

Step1: Identify the pattern

Looking at the table, we can try to find a relationship between \( x \) and \( y \). Let's list the points: \((-3, 0)\), \((-2, 6)\), \((-1, 2)\), \((0, 6)\), \((1, 0)\). Wait, maybe it's a quadratic function? Let's assume the general form of a quadratic function \( y = ax^2 + bx + c \).

Using the point \((0, 6)\), when \( x = 0 \), \( y = 6 \), so \( c = 6 \).

Now use \((-3, 0)\): \( 0 = a(-3)^2 + b(-3) + 6 \) → \( 9a - 3b + 6 = 0 \) → \( 3a - b = -2 \) (Equation 1)

Using \((1, 0)\): \( 0 = a(1)^2 + b(1) + 6 \) → \( a + b + 6 = 0 \) → \( a + b = -6 \) (Equation 2)

Now solve Equation 1 and Equation 2:

Add Equation 1 and Equation 2: \( (3a - b) + (a + b) = -2 + (-6) \) → \( 4a = -8 \) → \( a = -2 \)

Substitute \( a = -2 \) into Equation 2: \( -2 + b = -6 \) → \( b = -4 \)

So the quadratic function is \( y = -2x^2 - 4x + 6 \). Wait, but maybe we need to find \( x \) in terms of \( y \)? Wait, the problem says "Formula \( x = \) ______". Maybe it's a quadratic equation, so let's solve \( y = -2x^2 - 4x + 6 \) for \( x \).

First, rearrange: \( 2x^2 + 4x + (y - 6) = 0 \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 4 \), \( c = y - 6 \)

So \( x = \frac{-4 \pm \sqrt{16 - 8(y - 6)}}{4} = \frac{-4 \pm \sqrt{16 - 8y + 48}}{4} = \frac{-4 \pm \sqrt{64 - 8y}}{4} = \frac{-4 \pm 2\sqrt{16 - 2y}}{4} = \frac{-2 \pm \sqrt{16 - 2y}}{2} \)

Wait, but maybe looking at the symmetry. The vertex of the parabola \( y = -2x^2 - 4x + 6 \) is at \( x = -\frac{b}{2a} = -\frac{-4}{2(-2)} = -1 \). The roots are at \( x = -3 \) and \( x = 1 \) (since when \( y = 0 \), \( x = -3 \) and \( x = 1 \)). The axis of symmetry is \( x = -1 \).

Alternatively, maybe the table is symmetric around \( x = -1 \). Let's check the \( y \)-values: at \( x = -2 \) and \( x = 0 \), \( y = 6 \); at \( x = -1 \), \( y = 2 \); at \( x = -3 \) and \( x = 1 \), \( y = 0 \). So the vertex is at \( (-1, 2) \).

The standard form of a parabola is \( y = a(x - h)^2 + k \), where \( (h, k) = (-1, 2) \). So \( y = a(x + 1)^2 + 2 \). Using the point \( (1, 0) \): \( 0 = a(1 + 1)^2 + 2 \) → \( 4a + 2 = 0 \) → \( a = -\frac{1}{2} \). Wait, no, earlier we had \( a = -2 \). Wait, let's recalculate:

If \( y = a(x + 1)^2 + 2 \), using \( (1, 0) \): \( 0 = a(2)^2 + 2 \) → \( 4a = -2 \) → \( a = -\frac{1}{2} \). Then \( y = -\frac{1}{2}(x + 1)^2 + 2 \). Let's check \( x = 0 \): \( y = -\frac{1}{2}(1) + 2 = \frac{3}{2} \), but the table says \( y = 6 \) at \( x = 0 \). So my initial assumption of the points was wrong. Wait, maybe the table has some errors or I misread. Wait the original table:

Looking at the user's image:

x | y

-3 | 0

-2 | 6

-1 | 2

0 | 6

1 | 0

Ah! Wait, maybe it's a quadratic function symmetric about x = -1? Wait, no, the y-values at x=-2 and x=0 are both 6, x=-3 and x=1 are both 0, so the axis of symmetry is x = (-2 + 0)/2 = -1, or (-3 + 1)/2 = -1. So the vertex is at x = -1, y = 2. So the parabola opens downward? Wait, but when x=-1, y=2, which is the minimum? No, because at x=-2 and x=0, y=6 which is higher than 2, so it's a minimum? Wait, no, if the vertex is the minimum, then the parabola opens upward. Wait, let's recalculate the quadratic function correctly.

Using vertex form \( y = a(x - h)^2 + k \), h = -1, k = 2. So \( y = a(x + 1)^2 + 2 \).

Using point (0, 6): \( 6 = a(0 + 1)^2 + 2 \) → \( a + 2 = 6 \) → \( a = 4 \). So \( y = 4(x + 1)^2 + 2 \)? Wait, no, when x=0, y=6: 4(1) + 2 = 6, correct. x=-1: 4(0) + 2 = 2, correct. x=1: 4(4) + 2 = 18, but the table says x=1, y=0. So that's wrong. So clearly, my misreading of the table. Wait, maybe the y-values are different. Wait the user's handwritten table:

Looking again:

x: -3, y: 0

x: -2, y: 6

x: -1, y: 2

x: 0, y: 6

x: 1, y: 0

Ah! Now I see, it's symmetric about x = -1, and the roots are at x = -3 and x = 1 (since y=0 there), and the vertex at x=-1, y=2. So the quadratic function is \( y = -2(x + 3)(x - 1) \). Let's expand that: \( y = -2(x^2 + 2x - 3) = -2x^2 -4x + 6 \). Now check x=0: y=6, correct. x=-2: y=-2(4) -4(-2) +6 = -8 +8 +6=6, correct. x=-1: y=-2(1) -4(-1)+6= -2 +4 +6=8? Wait, no, the table says x=-1, y=2. Oh! So there's a mistake in my calculation. Wait, x=-1: y=-2(-1)^2 -4(-1) +6 = -2 +4 +6=8, but the table says y=2. So the table must have a typo, or I misread. Wait, maybe the y-value at x=-1 is 8? Or maybe the user made a mistake. Alternatively, maybe it's a different function.

Wait, maybe it's a absolute value function? No, the symmetry is quadratic. Alternatively, maybe the problem is to find x in terms of y, assuming it's a quadratic with roots at x=-3 and x=1, and vertex at x=-1, y=8 (if we correct the y-value at x=-1). Then the quadratic function is \( y = -2(x + 3)(x - 1) \), so solving for x:

\( y = -2(x^2 + 2x - 3) \)

\( y = -2x^2 -4x + 6 \)

\( 2x^2 + 4x + (y - 6) = 0 \)

Using quadratic formula:

\( x = \frac{-4 \pm \sqrt{16 - 8(y - 6)}}{4} = \frac{-4 \pm \sqrt{16 - 8y + 48}}{4} = \frac{-4 \pm \sqrt{64 - 8y}}{4} = \frac{-4 \pm 2\sqrt{16 - 2y}}{4} = \frac{-2 \pm \sqrt{16 - 2y}}{2} \)

But maybe the intended function is a quadratic with roots at x=-3 and x=1, and vertex at x=-1, y=8, so the formula for x in terms of y is \( x = -1 \pm \sqrt{\frac{8 - y}{2}} \) (since \( y = -2(x + 1)^2 + 8 \), so \( (x + 1)^2 = \frac{8 - y}{2} \), so \( x = -1 \pm \sqrt{\frac{8 - y}{2}} \))

Alternatively, maybe the table has a mistake, and the y-value at x=-1 is 8. Then the formula would be \( x = -1 \pm \sqrt{\frac{8 - y}{2}} \)

But given the table as is, with x=-1, y=2, it's inconsistent. Alternatively, maybe it's a different approach. Wait, the problem says "Formula x = ______", maybe it's a linear function? But the points are not linear. Wait, maybe the user wants to find the axis of symmetry? The axis of symmetry is x = -1, since the points are symmetric around x=-1 (x=-3 and x=1, x=-2 and x=0 have the same y-values). So the formula for x (the axis of symmetry) is x = -1? But that's only the axis. Alternatively, maybe the quadratic function's roots are x=-3 and x=1, so the average of the roots is x = (-3 + 1)/2 = -1, which is the axis of symmetry.

Alternatively, maybe the problem is to find x when y=0, but no, the formula is x=______.

Wait, maybe the table is for a quadratic function, and we need to express x in terms of y. Let's proceed with the quadratic function we derived earlier, even with the inconsistency at x=-1 (maybe a typo in the table). So the quadratic function is y = -2x² -4x +6. Solving for x:

2x² +4x + (y -6) = 0

x = [-4 ± √(16 - 8(y -6))]/4

x = [-4 ± √(16 -8y +48)]/4

x = [-4 ± √(64 -8y)]/4

x = [-4 ± 2√(16 -2y)]/4

x = [-2 ± √(16 -2y)]/2

Simplify:

x = -1 ± (√(16 - 2y))/2

Or factor out 2 from the square root:

√(16 - 2y) = √[2(8 - y)] = √2 √(8 - y)

So x = -1 ± (√2 √(8 - y))/2 = -1 ± √( (8 - y)/2 )

So the formula for x is \( x = -1 \pm \sqrt{\frac{8 - y}{2}} \)

Answer:

\( x = -1 \pm \sqrt{\frac{8 - y}{2}} \) (or equivalent form)