Parabolas & Non-linear Relationships

For $y = ax^2$, a negative value of $a$ makes the parabola open downward. Here $a = -2 < 0$, so it opens downward. (A positive $a$ opens upward.)

Set $x = 0$: $y = 0^2 - 5 = -5$. For $y = x^2 + c$ the y-intercept is always the constant $c$.

Set $h = 0$: $-5t^2 + 30t = 0$, so $-5t(t - 6) = 0$, giving $t = 0$ (thrown) or $t = 6$ (lands). It hits the ground at $t = 6$ seconds. (The x-intercepts of the parabola are the launch and landing times.)

If one number is $x$, the other is $12 - x$, so the product is $P = x(12 - x) = 12x - x^2$. This is a downward parabola with its maximum at the vertex $x = 6$, giving $P = 6 \times 6 = 36$. (The product is largest when the two numbers are equal.)

Set the $y$-values equal: $x^2 = x + 6$, so $x^2 - x - 6 = 0$ and $(x - 3)(x + 2) = 0$, giving $x = 3$ or $x = -2$. (A line can cut a parabola in two places.)

$y = (x - h)^2$ shifts horizontally and the sign inside reverses: $(x + 2)^2$ shifts left 2 units (vertex at $(-2, 0)$). (Right 2 would be $(x - 2)^2$.)

Width depends on $|a|$: the larger $|a|$ is, the narrower the parabola. The largest here is $|a| = 2$, so $y = 2x^2$ is the narrowest. ($y = 0.1x^2$ has the smallest $|a|$, so it is the widest.)

A parabola is symmetric about its axis, so two points with the same $y$-value are mirror images across $x = 2$. The point $(5, 7)$ is 3 units right of the axis, so its mirror is 3 units left: $x = 2 - 3 = -1$, giving $(-1, 7)$.

In vertex form $y = (x - h)^2 + k$, the vertex is $(h, k)$. For $(x - 2)^2 + 5$: $h = 2$, $k = 5$, so the vertex is $(2, 5)$. ($(-2, 5)$ did not reverse the inside sign.)

The negative sign flips the parabola upside down, a reflection in the x-axis (so it now opens downward). ($y = x^2$ is symmetric about the y-axis, so reflecting it in the y-axis leaves it unchanged.)

The equation is in vertex form $h = a(x - p)^2 + k$, and since $a = -0.5 < 0$ the parabola opens downward, so the vertex is the maximum. The maximum height is $k = 8$ m (reached at $x = 4$). ($4$ m is the horizontal position of the peak, not the height.)

Factorise: $x^2 - 2x - 3 = (x - 3)(x + 1) = 0$, so $x = 3$ or $x = -1$. ($x = -3 \text{ and } x = 1$ reversed the signs.)

In $y = (x - h)^2 + k$ the vertex is $(h, k)$. Here $(x + 1)^2 - 4$ gives $h = -1$ (sign reverses) and $k = -4$, so the vertex is $(-1, -4)$.

The axis of symmetry passes midway between the x-intercepts: $x = \dfrac{-2 + 4}{2} = 1$. (A parabola is symmetric, so its axis is exactly halfway between the roots.)

Set $x = 0$: $y = (0 - 2)(0 + 5) = (-2)(5) = -10$. (The y-intercept is the product of the two constants.)

$x^2$ is at least 0, so the smallest value of $y = x^2 + 4$ is $0 + 4 = 4$, reached at $x = 0$ (the vertex). (Since it opens upward, it has a minimum, not a maximum.)

A parabola $y = ax^2$ opens downward exactly when the coefficient $a$ is negative, i.e. $a < 0$. (Any negative value works, not just $-1$.)

Substitute $x = 2$: $y = 2^2 - 1 = 3$, so $(2, 3)$ is on the curve. (Checking $(0, 1)$: $0 - 1 = -1 \ne 1$, so it is not on the parabola.)

Factorise: $x^2 - 6x + 5 = (x - 1)(x - 5) = 0$, so $x = 1$ or $x = 5$ (two numbers that multiply to 5 and add to $-6$ are $-1$ and $-5$).

$A = x(20 - x) = 20x - x^2$ is a downward parabola with its maximum at the vertex $x = 10$. Then $A = 10 \times 10 = 100$ m² (a square pen). (The area is greatest when the rectangle is a square.)

A parabola comes from a quadratic (an $x^2$ term). $y = 3x + 2$ has no $x^2$ term, so it is a straight line, not a parabola. All the others contain $x^2$.

The factored form is $y = x(x - 6) = x^2 - 6x$, so $b = -6$. (The roots $0$ and $6$ give the factors $x$ and $(x - 6)$.)

Substitute $t = 1$: $h = -5(1)^2 + 20(1) = -5 + 20 = 15$ m. (Squaring the 1 first, then multiplying.)

Since $a = 1 > 0$, the parabola opens upward, so the vertex is a minimum. The $+2$ lifts the vertex to $(0, 2)$.

The $2$ makes it narrower (larger $|a|$); $(x - 1)^2$ shifts it right 1; $+3$ shifts it up 3. So the vertex is at $(1, 3)$ and the curve is narrower than $y = x^2$.