- In the previous article, we covered quadratic expressions.
- Now, we can talk about quadratic functions.
- A "U-shaped" (or upside-down U-shaped) relationships such as projectiles, arches, and optimization problems.
Quadratic function
A function whose graph is a parabola and can be written in the form $y=ax^2+bx+c$ where $a\neq 0$.
Key Features Of Any Quadratic Graph
When you sketch or analyze a quadratic, you nearly always focus on three main characteristics:three main characteristics:
- Intercepts (where the graph crosses the axes)
- Vertex (the turning point)
- Axis of symmetry (the vertical line through the vertex)
x-Intercepts (zeros)
The $x$-coordinates where the graph crosses the $x$-axis (where $y=0$). They are also called the zeros (or roots) of the function.
y-Intercept
The point where the graph crosses the $y$-axis (where $x=0$).
Vertex
The turning point of the parabola, with coordinates $(x_v, y_v)$.
Axis of symmetry
The vertical line that divides the parabola into two mirror-image halves. Its equation is $x=x_v$.
How The Parameters $a$, $b$, And $c$ Affect The Parabola
- In standard form, $$y=ax^2+bx+c,\quad a\neq 0$$
- $a$ controls opening and steepness:
- If $a>0$, the parabola opens upward (minimum point).
- If $a<0$, the parabola opens downward (maximum point).
- The larger $|a|$ is, the narrower (steeper) the parabola; if $0<|a|<1$, it is wider.
- $c$ is the y-intercept (because when $x=0$, $y=c$).
- $b$ influences where the vertex sits horizontally, and therefore the axis of symmetry.
The parameter $b$ does not have a simple "one-step" interpretation like $c$, but it is crucial because it shifts the axis of symmetry and vertex left or right.
Intercepts: Fast Ways To Find Them
Finding The y-Intercept
- Set $x=0$:
- So the y-intercept is $(0,c)$.
Finding x-Intercepts By Solving $y=0$
- To find x-intercepts, set $y=0$, set $y=0$: $$ax^2+bx+c=0$$
- How you solve depends on the form of the quadratic.
- If it factorizes, you can use the zero-product property.
- If it does not factorize nicely, you can still find intercepts using a method like the quadratic formula (often taught later), or by completing the square / technology.
- A quadratic can fail to factorize over the integers and still have real x-intercepts.
- "Not factorisable (nicely)" does not automatically mean "no intercepts".
Factorized Form Makes Zeros Visible
- A quadratic written in factorized form looks like $$y=a(x-r_1)(x-r_2)$$ where $r_1$ and $r_2$ are the zeros.
- Setting $y=0$ gives
- So the x-intercepts are $x=r_1$ and $x=r_2$ (points $(r_1,0)$ and $(r_2,0)$).
- If $y=(x-5)(x+1)$, then $y=0$ when $x-5=0$ or $x+1=0$.
- So the x-intercepts are $x=5$ and $x=-1$.
The Vertex Lies Halfway Between The x-Intercepts
- If a parabola has two x-intercepts $xx-intercepts $x_1$ and $x_2$, then the axis of symmetry passes exactly halfway between them.
- That means the x-coordinate of the vertex is the average: $$x_v=\frac{x1+x_2}{2}$$
- This is a powerful checking tool when you can see or calculate the intercepts.checking tool when you can see or calculate the intercepts.
- Think of the axis of symmetry as a "mirror line".
- The two x-intercepts are mirror images across that line, so the mirror must be exactly halfway between them.
- Students sometimes average the points incorrectly by averaging the y-values too.
- For the vertex x-coordinate, you average only the x-intercepts: $$x_v=\frac{x_1+x_2}{2}$$
The Vertex x-Coordinate In Standard Form Is $x
- For any quadratic in standard form $y=ax^2+bx+c$, the vertex x-coordinate is $$x_v=-\frac{b}{2a}$$
- This formula works whether $a=1$, $a=-1$, or any other nonzero value.
Why This Matches "Halfway Between The Zeros"
- When a quadratic is factorized as $y=a(x-r_1)(x-r_2)$, expanding gives $$y=a\left(x^2-(r_1+r_2)x+r_1r_2\right)$$
- So in standard form, $b=-a(r_1+r_2)$.
- Then: $$-\frac{b}{2a}=-\frac{-a(r_1+r_2)}{2a}=\frac{r_1+r_2}{2}$$ which is exactly the average of the x-intercepts.
Finding The Vertex y-Coordinate Once You Know $x_v$
Once you have $x_v$, substitute it back into the function: $$y_v=f(x_v)$$
So the vertex is $(x_v, f(x_v))$.
A reliable routine is: (1) compute $x_v=-\frac{b}{2a}$, then (2) plug it into $y=ax^2+bx+c$ to get $y_v$.
Find the vertex of $y=x^2-4x-5$.
Solution
- $a=1$, $b=-4$ so $$x_v=-\frac{b}{2a}=-\frac{-4}{2\cdot 1}=2$$
- Substitute $x=2$: $$y_v=2^2-4(2)-5=4-8-5=-9$$
- So the vertex is $(2,-9)$.
Vertex Form Shows The Vertex Immediately
A quadratic in vertex form is $$y=a(x-h)^2+k,\quad a\neq 0$$
Vertex form
A form of a quadratic function $y=a(x-h)^2+k$ where $(h,k)$ is the vertex.
From this form:
- The vertex is $(h,k)$.
- The axis of symmetry is $x=h$.
Translation Up And Down
Adding or subtracting a constant changes $k$:
- $y=(x-2)^2+4$ is the graph of $y=(x-2)^2$ shifted up 4 units, vertex $(2,4)$, and it has no x-intercepts.
- $y=(x-2)^2-3$ is shifted down 3 units, vertex $(2,-3)$, and it has two x-intercepts.
- Whether a parabola has 0, 1, or 2 x-intercepts depends on whether it meets the x-axis.
- In vertex form, you can often see this quickly by comparing $k$ to 0 and considering whether it opens up or down.
When There Is Only One x-Intercept (A Repeated Root)
- Sometimes a parabola touches the x-axis at exactly one point, at the vertex.
- This happens when the quadratic has a repeated factor.
- In general: $$y=a(x-h)^2$$
- Vertex is $(h,0)$
- There is exactly one unique x-intercept, $x=h$, but it is a repeated zero.
Repeated root (double zero)
A zero where the factor repeats, for example $(x-h)^2$. The graph touches the x-axis at $x=h$ but does not cross it.
Three Equivalent Forms And When Each Is Best
Quadratics are commonly represented in three forms:
- Standard form: $$y=ax^2+bx+c$$
- Best for spotting the y-intercept ($c$) and using algebraic methods like completing the square.
- Factorized form: $$y=a(x-r_1)(x-r_2) \text{ or } y=a(x-h)^2$$
- Best for finding x-intercepts quickly.
- Vertex form: $$y=a(x-h)^2+k$$
- Best for reading the vertex and axis of symmetry immediately, and understanding translations.
- In problems where you are asked for intercepts, try factorized form first.
- If you are asked for maximum/minimum value, use vertex form or compute $(x_v,y_v)$ from standard form.
For each function below, state (i) whether it opens up or down, (ii) the y-intercept, and (iii) the vertex x-coordinate.
- $y=-x^2+2x+10$
- $y=2x^2+4x-1$
- $y=\tfrac12 x^2+2x-5$
