Functions and Graphs

$y = ax^2 + k$ has its vertex at $(0, k)$. For that vertex to sit on the positive $y$-axis, its height must be positive, so $k > 0$. ($a$ controls only the width and direction, not the vertex height.)

The $x$-coordinate increases by $4 - 1 = 3$ (right $3$) and the $y$-coordinate decreases by $2 - 0 = 2$ (down $2$). Every point of the graph moves right $3$ and down $2$.

The original vertex is $(2, 3)$. Reflecting in the $x$-axis negates every $y$-coordinate, leaving $x$ unchanged, so the vertex maps to $(2, -3)$. ($(-2, 3)$ reflected in the $y$-axis instead.)

Set $y = 0$ and factorise: $x^2 - 6x + 8 = (x - 2)(x - 4) = 0$, so $x = 2$ or $x = 4$. ($x = -2 \text{ and } x = -4$ forgot the sign-flip from the brackets to the solutions.)

When $0 < |a| < 1$ the parabola is wider (flatter): at $x = 2$, $y = \dfrac{1}{2}(4) = 2$ versus $4$ for $y = x^2$. ("Narrower" is the common misconception; $|a| > 1$ is what narrows it.)

The gradient is $m = \dfrac{5 - (-1)}{2 - 0} = \dfrac{6}{2} = 3$, and the line cuts the $y$-axis at $(0, -1)$, so the intercept is $-1$. Thus $y = 3x - 1$.

Perpendicular gradients multiply to $-1$, so the new gradient is $-\dfrac{1}{2}$ (the negative reciprocal of $2$). Through $(0, 3)$ the intercept is $3$, giving $y = -\dfrac{1}{2}x + 3$.

Read the changes in turn: $(x - 1)^2$ shifts right $1$; the factor $2$ stretches it vertically (narrower) by $2$; $+3$ shifts up $3$. So: right $1$, stretch $\times 2$, up $3$.

The parabola opens upward with vertex $(2, -1)$ and crosses the $x$-axis at $1$ and $3$. Vertex form $y = (x - 2)^2 - 1$ matches. ($(x + 2)^2 - 1$ has vertex $(-2, -1)$; the $-(\dots)$ option opens downward; $+1$ lifts the vertex above the axis, giving no real roots.)

Three transformations: $(x + 1)^2$ shifts left $1$, $+4$ shifts up $4$, and the leading negative reflects in the x-axis. So: left $1$, up $4$, reflected.

As $x \to \infty$, $\dfrac{1}{x}$ gets smaller and smaller but stays positive, approaching 0 without ever reaching it. The $x$-axis ($y = 0$) is a horizontal asymptote, so $y$ never actually equals 0.

The $x$-intercept is where $y = 0$: $0 = 2x + 6$ gives $x = -3$. The $y$-intercept is $6$. The right-angled triangle has base $3$ and height $6$, so its area is $\dfrac{1}{2} \times 3 \times 6 = 9$ units$^2$.

On $y = \dfrac{k}{x}$ the product $xy = k$ is constant, so $k = 3 \times 4 = 12$. Then $y = \dfrac{12}{6} = 2$. ($12$ stopped at $k$.)

For $y = \dfrac{k}{x}$, negative $k$ puts the branches in quadrants 2 and 4 (where $x$ and $y$ have opposite signs). (Positive $k$ gives quadrants 1 and 3.)

The parabola opens downward with vertex $(-1, 4)$ and crosses the $x$-axis at $-3$ and $1$. Vertex form $y = -(x + 1)^2 + 4$ matches. ($(x + 1)^2 + 4$ opens upward; $-(x - 1)^2 + 4$ has vertex $(1, 4)$; the others put the vertex on or below the axis.)

The centre is the midpoint of the diameter: $\left(\dfrac{1 + 5}{2}, \dfrac{2 + 8}{2}\right) = (3, 5)$. ($(6, 10)$ added the coordinates without halving.)

The radius is the distance from the centre to the point: $r = \sqrt{(5 - 2)^2 + (3 - (-1))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. ($\sqrt{7}$ subtracted the squares instead of adding.)

The right side is $r^2$, so $r = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \approx 4.47$. ($\sqrt{20}$ is correct but not simplified; $20$ is $r^2$, not $r$.)

Set them equal: $x^2 = x + 2$, so $x^2 - x - 2 = 0$ and $(x - 2)(x + 1) = 0$, giving $x = 2$ or $x = -1$. The points are $(2, 4)$ and $(-1, 1)$.

Substitute: $3^2 + 4^2 = 9 + 16 = 25$. Since $25 > 16$, the point is outside the circle (it is $\sqrt{25} = 5$ from the origin, but the radius is only $4$).

$y = (x^2 + 4x + 4) + 5 - 4 = (x + 2)^2 + 1$, so the vertex is $(-2, 1)$. (Check: $y(-2) = 4 - 8 + 5 = 1$.)

$(x - 2)^2 + (y - (-3))^2 = 4^2$, i.e. $(x - 2)^2 + (y + 3)^2 = 16$. (Use $r^2 = 16$, not $r = 4$.)

Substitute $x = 0$: $y = (-3)^2 + 1 = 9 + 1 = 10$, so $(0, 10)$. ($(-3)^2 = 9$, not $-9$; $(3, 1)$ is the vertex.)

The denominator is zero when $x - 2 = 0$, i.e. $x = 2$, which is undefined and so excluded. (There is a vertical asymptote at $x = 2$.)

Vertex $(2, 0)$ gives $y = a(x - 2)^2$; through $(0, 4)$: $4 = a(4)$, so $a = 1$ and $y = (x - 2)^2$. ($y = -(x - 2)^2$ opens downward.)