Quadratic Functions
A quadratic function has the form
\(f(x)=ax^2+bx+c\) with \(a\neq 0\).
Its graph is a parabola whose shape and position are controlled by \(a,\,b,\,c\).
Three Forms at a Glance
- Standard form: \(f(x)=ax^2+bx+c\).
- Vertex form: \(f(x)=a(x-h)^2+k\) (vertex \((h,k)\), axis \(x=h\)).
- Factored form: \(f(x)=a(x-p)(x-q)\) (zeros \(x=p,q\), axis \(x=\displaystyle\frac{p+q}{2}\)).
Core Features
- Opens: up if \(a>0\), down if \(a<0\).
- Width: narrower if \(|a|>1\), wider if \(|a|<1\).
- Axis of symmetry: \(x=-\displaystyle\frac{b}{2a}\).
- Vertex: \(\left(-\displaystyle\frac{b}{2a},\,f\!\left(-\displaystyle\frac{b}{2a}\right)\right)\).
- y-intercept: \((0,c)\).
Minimum / Maximum, Domain & Range
- If \(a>0\) (opens up): vertex value \(f\!\left(-\displaystyle\frac{b}{2a}\right)\) is a minimum; domain \((-\infty,\infty)\); range \(\left[y_{\min},\infty\right)\) with \(y_{\min}=f\!\left(-\displaystyle\frac{b}{2a}\right)\); decreasing left of the vertex, increasing right.
- If \(a<0\) (opens down): vertex value \(f\!\left(-\displaystyle\frac{b}{2a}\right)\) is a maximum; domain \((-\infty,\infty)\); range \(\left(-\infty,y_{\max}\right]\) with \(y_{\max}=f\!\left(-\displaystyle\frac{b}{2a}\right)\); increasing left of the vertex, decreasing right.
Converting Forms (Completing the Square)
Starting from \(f(x)=ax^2+bx+c\) with \(a\neq 0\),
\(\displaystyle f(x)=a\!\left(x+\displaystyle\frac{b}{2a}\right)^2+\left(c-\displaystyle\frac{b^2}{4a}\right) \;=\;a(x-h)^2+k, \)
where
\(h=-\displaystyle\frac{b}{2a}\) and \(k=c-\displaystyle\frac{b^2}{4a}\).
Roots, Quadratic Formula & Discriminant
To solve \(ax^2+bx+c=0\): \(\displaystyle x=\frac{-b\pm\sqrt{\,b^2-4ac\,}}{2a}\).
Discriminant: \(D=b^2-4ac\).
- \(D>0\): two real, distinct zeros → two \(x\)-intercepts.
- \(D=0\): one real, repeated zero (touches the axis at the vertex).
- \(D<0\): no real zeros (parabola lies entirely above or below the \(x\)-axis).
Vieta’s Relations
For real/complex roots \(r_1,\,r_2\),
\(r_1+r_2=-\displaystyle\frac{b}{a}\)
and
\(r_1r_2=\displaystyle\frac{c}{a}\).
Standard vs. Vertex vs. Factored
| Form | Equation | Axis of Symmetry | Vertex | $x$-Intercepts | $y$-Intercept | Best For |
|---|---|---|---|---|---|---|
| Standard | \(f(x)=ax^2+bx+c\) | \(x=-\displaystyle\frac{b}{2a}\) | \(\left(-\displaystyle\frac{b}{2a},\,f\!\left(-\displaystyle\frac{b}{2a}\right)\right)\) | Use quadratic formula / factoring if possible | \((0,c)\) | Reading coefficients; applying formulae |
| Vertex | \(f(x)=a(x-h)^2+k\) | \(x=h\) | \((h,k)\) | Solve \(a(x-h)^2+k=0\) | \((0,\,a h^2+k)\) | Graphing min/max, range, transformations |
| Factored | \(f(x)=a(x-p)(x-q)\) | \(x=\displaystyle\frac{p+q}{2}\) | \(\left(\displaystyle\frac{p+q}{2},\,f\!\left(\displaystyle\frac{p+q}{2}\right)\right)\) | \((p,0)\) and \((q,0)\) | \(\big(0,\,a\,pq\big)\) | Finding zeros and axis quickly |
Quick Graphing Checklist
- Identify \(a\) (opening & width): up/down, narrower/wider.
- Find axis \(x=-\displaystyle\frac{b}{2a}\) (or \(x=h\) in vertex form).
- Compute vertex value \(f\!\left(-\displaystyle\frac{b}{2a}\right)\) (or read \(k\)).
- Get intercepts: \(y\)-intercept \((0,c)\); \(x\)-intercepts by solving \(f(x)=0\).
- Plot symmetric points about the axis and sketch the parabola.
Common Pitfalls
- Forgetting the negative in \(x=-\displaystyle\frac{b}{2a}\).
- Completing the square without factoring out \(a\) first when \(a\neq 1\).
- Assuming intercepts exist even when \(D<0\).
Example 1
Graph \(f(x) = -2(x+3)^2 + 4\). Label the vertex and axis of symmetry.
Solution:
Step 1: Identify the constants. Here \(a=-2\), \(h=-3\), and \(k=4\).
Step 2: Plot the vertex \((h,k)=(-3,\,4)\) and draw the axis of symmetry \(x=-3\).
Step 3: Evaluate \(f(x)\) at two points to guide the graph:
- \(f(-2) = -2(-2+3)^2+4 = 2\)
- \(f(-1) = -2(-1+3)^2+4 = -4\)
Plot the points \((-2,\,2)\) and \((-1,\,-4)\) along with their reflections across the axis \(x=-3\).
Step 4: Sketch a downward-opening parabola through the points, with vertex at \((-3,\,4)\).
Notice that quadratic functions can be written in standard form as well:
\(\begin{align*} f(x) &= a(x-h)^2 + k && \text{(vertex form)} \\ &= a(x^2 – 2hx + h^2) + k && \text{(expand)} \\ &= ax^2 – 2ahx + (ah^2 + k) && \text{(simplify)} \\ &= ax^2 + bx + c && \text{(standard form)} \end{align*}\)
Here:
- \(a = a\) (same in both forms).
- \(b = -2ah\) (so \(h=-\displaystyle\frac{b}{2a}\), axis of symmetry \(x=-\displaystyle\frac{b}{2a}\)).
- \(c = ah^2 + k\) (so \(c\) is the y-intercept).
Example 2
Graph \(f(x) = 3x^2 – 6x + 1\). Label the vertex and axis of symmetry.
Solution:
Step 1: Identify the coefficients: \(a=3\), \(b=-6\), and \(c=1\). Since \(a>0\), the parabola opens upward.
Step 2: Find the vertex.
\(x = -\displaystyle\frac{b}{2a} = -\displaystyle\frac{-6}{2(3)} = 1\)
Then substitute \(x=1\) to find the \(y\)-coordinate:
\(f(1) = 3(1)^2 – 6(1) + 1 = -2\)
So, the vertex is \((1,-2)\).
Step 3: Draw the axis of symmetry \(x=1\).
Step 4: Identify the \(y\)-intercept \(c=1\). So the point \((0,1)\) is on the parabola. Reflect this point across \(x=1\) to get \((2,1)\).
Step 5: Choose another \(x\)-value, such as \(x=3\).
\(f(3) = 3(3)^2 – 6(3) + 1 = 10\)
So \((3,10)\) is on the parabola. Its reflection across \(x=1\) is \((-1,10)\).
Step 6: Plot all the points and sketch the parabola, symmetric about \(x=1\).
Example 3
Find the minimum or maximum value of \(f(x)=\displaystyle\frac{1}{2}x^2 – 2x – 1\).
Describe the domain and range, and determine where the function is increasing and decreasing.
Solution:
Step 1: Identify the coefficients: \(a=\displaystyle\frac{1}{2}\), \(b=-2\), and \(c=-1\). Since \(a>0\), the parabola opens upward and has a minimum value.
Step 2: Find the vertex.
\(x = -\displaystyle\frac{b}{2a} = -\displaystyle\frac{-2}{2\left(\displaystyle\frac{1}{2}\right)} = 2\)
\(f(2) = \displaystyle\frac{1}{2}(2)^2 – 2(2) – 1 = -3\)
So, the vertex is \((2, -3)\), and the minimum value is \(-3\).
Step 3: State the domain and range.
- Domain: all real numbers (\(-\infty < x < \infty\)).
- Range: \(y \geq -3\).
Step 4: Determine intervals of increase and decrease.
- The function is decreasing for \(x < 2\).
- The function is increasing for \(x > 2\).
Example 4
Graph \(f(x) = -2(x+3)(x-1)\).
Label the \(x\)-intercepts, vertex, and axis of symmetry.
Solution:
Step 1: Identify the \(x\)-intercepts. Since \(f(x) = -2(x+3)(x-1)\), the roots are \(p=-3\) and \(q=1\).
So the \(x\)-intercepts are \((-3,0)\) and \((1,0)\).
Step 2: Find the vertex.
\(x = \displaystyle\frac{p+q}{2} = \displaystyle\frac{-3+1}{2} = -1\)
\(f(-1) = -2((-1)+3)((-1)-1) = -2(2)(-2) = 8\)
So the axis of symmetry is \(x=-1\) and the vertex is \((-1,8)\).
Step 3: Draw the parabola through the vertex and the intercepts.
The parabola opens downward (since \(a=-2<0\)).
Practice
Graph each quadratic function. Label the vertex, axis of symmetry, and intercepts as required. State the minimum/maximum value, domain, and range where appropriate.
1. \(f(x) = -3(x+1)^2\). Label the vertex and axis of symmetry.
2. \(g(x) = 2(x-2)^2 + 5\). Label the vertex and axis of symmetry.
3. \(h(x) = x^2 + 2x – 1\). Label the vertex and axis of symmetry.
4. \(p(x) = -2x^2 – 8x + 1\). Label the vertex and axis of symmetry.
5. (a) \(f(x) = 4x^2 + 16x – 3\) (b) \(h(x) = -x^2 + 5x + 9\).
For each: Find the minimum or maximum value, describe the domain and range, and determine where the function is increasing and decreasing.
6. \(f(x) = -(x+1)(x+5)\). Label the \(x\)-intercepts, vertex, and axis of symmetry.
7. \(g(x) = \displaystyle\frac{1}{4}(x-6)(x-2)\). Label the \(x\)-intercepts, vertex, and axis of symmetry.
8. The quadratic equation \(2x^2 – 5x + 3 = 0\) has roots \(r_1\) and \(r_2\).
(a) Use Vieta’s formulas to find \(r_1 + r_2\) and \(r_1 r_2\).
(b) Without solving the quadratic, determine the value of \(r_1^2 + r_2^2\).
