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maths1-week3
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Now · Week map
Parabolas link algebra (polynomial degree 2), graphs (vertex and symmetry), and root-finding.
Week map
Standard form ax2+bx+c → graph shape → factoring and roots → quadratic formula → vertex and axis → discriminant → parameter problems.
Quadratic notation
- f(x)=ax2+bx+c → standard quadratic → a=0 → f(x)=2x2−4x+1.
- a → leading coefficient → sign sets cup up (a>0) or down (a<0).
- Root / zero → input where f(x)=0 → solve ax2+bx+c=0.
- h=−2ab → x-coordinate of vertex → axis of symmetry vertical line x=h.
- D=b2−4ac → discriminant → classifies real roots.
Vertex form
f(x)=a(x−h)2+k → vertex (h,k) → graph shifts h horizontal, k vertical.
Mini-example: f(x)=3(x−2)2−5 has vertex (2,−5), opens upward (a=3>0).
Roots and the discriminant
| D | Real roots |
|---|---|
| D>0 | two distinct |
| D=0 | one repeated (touch axis) |
| D<0 | none (no horizontal crossing) |
Quadratic formula: x=2a−b±D when solving ax2+bx+c=0.
Mini-example: x2−5x+6=0. Factor (x−2)(x−3)=0 → roots 2,3. Check D=25−24=1>0.
Factoring strategy (monic a=1)
Find integers p,q with pq=c and p+q=b for x2+bx+c.
x2+x−12: need product −12, sum 1 → 4 and −3 → (x+4)(x−3).
If no integer pair works, use formula or completing the square.
Completing the square sketch
x2+6x+1=(x2+6x+9)−9+1=(x+3)2−8. Vertex (−3,−8).
Graph features
- Axis of symmetry: x=h.
- y-intercept: f(0)=c.
- Direction: min value at vertex if a>0; max if a<0.
- Width: larger ∣a∣ → narrower parabola.
Pattern families
Easy — Factor and solve
- Monic quadratics with integer roots.
- Read vertex from (x−h)2+k when given.
- State y-intercept from c.
Medium — Vertex and transformation
- Convert standard to vertex form by completing square or formula for h,k.
- Given roots, write quadratic in factored form a(x−r1)(x−r2).
- Use symmetry: roots equidistant from axis at h.
Hard — Discriminant parameters
- Find values of k so x2+kx+9=0 has exactly one real root → D=0.
- Inequalities on D for “no real roots” or “two real roots”.
- Word constraints linking vertex height to domain (e.g. minimum cost).
Worked mini-examples
Example 1 — Factoring. 2x2−8x+6=0. Divide 2: x2−4x+3=0 → (x−1)(x−3)=0 → x=1,3.
Example 2 — Vertex. f(x)=−x2+4x+1. h=−4/(2⋅−1)=2. f(2)=−4+8+1=5. Vertex (2,5), max (since a=−1).
Example 3 — Discriminant. 3x2−2x+k=0 has no real roots when D=4−12k<0 → k>31.
Example 4 — From roots. Roots −2 and 5: f(x)=a(x+2)(x−5). If a=1, f(x)=x2−3x−10. c=−10 is y-intercept.
Traps
- Sign error in h=−b/(2a) when b negative.
- Vertex (h,k) confused with roots (zeros on x-axis).
- Forgetting a when using formula: divide by 2a, not 2.
- D=0 means one repeated root, not “no roots”.
- Opening direction from sign of a, not c.
Diagnostic (try yourself)
-
Solve x2+7x+10=0 by factoring.
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Find the vertex and axis of symmetry of f(x)=2x2−12x+5.
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For what values of m does x2+mx+16=0 have exactly one real solution?
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A parabola has roots at x=1 and x=9 and passes through (0,18). Write f(x) in standard form.
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Without fully solving, how many real roots does 2x2+3x−7=0 have? Use the discriminant.