The parabola y = a(x + p)^2 + q

For $y=a(x+p{)}^2+q$:

• $a$ controls shape and direction. $a>0$ opens upwards (minimum); $a<0$ opens downwards (maximum). A larger $|a|$ gives a narrower graph.
• $p$ shifts the graph horizontally: $p$ units left if $p>0$, right if $p<0$.
• $q$ shifts the graph vertically: $q$ units up if $q>0$.

Example. $y=2(x-1{)}^2-8$ is $y=2x^2$ shifted 1 unit right and 8 units down. It opens upwards and is narrower than $y=x^2$.

For $y=a(x+p{)}^2+q$:

• Turning point: $(-p;q)$
• Axis of symmetry: $x=-p$
• Range: $y\ge q$ if $a>0$, or $y\le q$ if $a<0$
• $y$-intercept: let $x=0$
• $x$-intercepts: let $y=0$ and solve

Example. $y=2(x-1{)}^2-8$

• Turning point $(1;-8)$, axis $x=1$, range $y\ge -8$
• $y$-intercept: $2(1)-8=-6$
• $x$-intercepts: $(x-1{)}^2=4$, so $x-1=\pm 2$, giving $x=3$ or $x=-1$

The $x$-intercepts are symmetrical about the axis: $-1$ and $3$ are both 2 units from $x=1$.

1. Decide the shape from the sign of $a$.
2. Plot the turning point.
3. Plot the $y$-intercept and its mirror image in the axis of symmetry.
4. Plot the $x$-intercepts, if there are any.
5. Draw a smooth curve and label every point with coordinates.

If the equation is given as $y=a{x}^2+bx+c$, find the turning point with $x=-\frac{b}{2a}$, or complete the square.

Example. $y=x^2-6x+5=(x-3{)}^2-4$. Turning point $(3;-4)$, $y$-intercept 5, $x$-intercepts 1 and 5.

Turning point given. Use $y=a(x+p{)}^2+q$, substitute the turning point, then use another point to find $a$.

Example. Turning point $(-2;5)$, passing through $(0;1)$:

$1=a(0+2{)}^2+5\Rightarrow 4a=-4\Rightarrow a=-1$

$y=-(x+2{)}^2+5=-x^2-4x+1$

$x$-intercepts given. Use $y=a(x-x_1)(x-x_2)$ and another point to find $a$.

Check your equation with a point you did not use.