Here you will find original math practice problems aligned with the IB Math SL analysis and approaches curriculum. Visit this site directly at hunkim.com/sl
IB Math SL Overview
- Lines
- Quadratic Functions
- Exponents and Logarithms
- Rational Functions
- Binomial Theorem
- Trigonometry
- Statistics
- Calculus
IB Math SL Chapter 1: Lines
- Equations of straight lines
- Parallel and perpendicular lines
Warm-Up
- Points A(1,2) and B(4,-7).
Equation in the form y=mx+b?
Solutiony=-3x+5 - L_1: 3y-2x=4
L_2 is perpendicular to L_1 and has an x-intercept of 4. Find L_2 in the form y=mx+b.
Solutiony=-\frac{3}{2}x+6 - f(x)=e^x. The slope at x=2 is e^2.
- Find the equation of the tangent line at x=2 in the form f(x)=mx+b.
Solutionf(x)=e^2 x-e^2 - y-intercept of the tangent line?
Solution-e^2
- Find the equation of the tangent line at x=2 in the form f(x)=mx+b.
IB Math SL Chapter 2: Quadratics
- Solving quadratic equations and inequalities
- Discriminant
- Domain, range, inverse
- Composite functions, identity
- Graphing and transformation functions
Warm-Up
- Sketch y=x^2-4
Solution - Sketch y=(x-2)^2
Solution - f(x)=(x+1)^2-4
- Sketch
Solution - Coordinates of the vertex?
Solution(-1,-4) - Domain?
Solutionx\in\mathbb{R} - Range?
Solutiony\geq -4 - Evaluate f(2)
Solution5 - Given x>2 find the range of f(x)
Solutiony>5
- Sketch
- y=-2(x-3)^2+8
- Sketch
Solution - Write in the form y=ax^2+bx+c
Solutiony=-2x^2+12x-10 - Value of the discriminant?
Solution64 - Find the domain in which f(x)>6
Solution2<x<4
- Sketch
- f(x)=2(x-2)(x+4)
- Intercepts?
Solution2, -4 - Equation of the line of symmetry?
Solutionx=-1 - Coordinates of the vertex?
SolutionV(-1,-18)
- Intercepts?
- What is the equation of the quadratic below in the form:
y=ax^2+bx+c?
Solutiony=2(x-2)^2-2 - Solve y=2x^2-5x+3
- By factoring
Solution1, \frac{3}{2} - By using the quadratic formula x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
Solution1, \frac{3}{2} - By completing the square
Solution1, \frac{3}{2}
- By factoring
- Solve h(t)=-3t^2-9t
Solution0, -3 - Solve points of intersections of the simultaneous equations:
y=2x^2-x and y-2=2x
Solution(-0.5,1) and (2,6) - f(x)=x^2-4x+3. Describe the transformation to move this parabola’s vertex to the origin.
SolutionTranslate 2 units left and 1 unit up - Write in vertex form: y=\frac{2}{3}x^2-\frac{x}{2}+1
Solutiony=\frac{2}{3}\left(x-\frac{3}{8}\right)^2+\frac{29}{32} - Given f(x)=(x-2)^2+4 what are the coordinates of the maximum of \frac{1}{f(x)}?
Solution\left(2,\frac{1}{4}\right) - f(x)=x^2+2x-3. How far away is the point (2,0) from the vertex?
Solution5 units - f(x)=2px^2+(p-8)x+\frac{3}{5}p-1
- Show that the discriminant is -\frac{19}{5}p^2-8p+64
- Find the values of p so that f(x)=0 has two equal roots
Solutionp\approx -5.29, 3.18
- Show that the discriminant is -\frac{19}{5}p^2-8p+64
- Solve:
- x^2\leq 9
Solution-3\leq x\leq 3 - x^2>4
Solutionx<-2 or x>2 - x^2<5
Solution-\sqrt{5}<x<\sqrt{5} - (x-2)^2<25
Solution-3<x<7 - (2x+1)^2\geq 7
Solutionx\leq \frac{\sqrt{7}-1}{2} or x\geq \frac{\sqrt{2}-1}{2}
- x^2\leq 9
- g(x)=x^2+bx+11. P(-1,8) is on the graph of g. Find b.
Solution4 - Find k for which the equations 2x^2+6x+k=0 has repeated roots.
Solution\frac{9}{2} - f(x)=(k-2)x^2+x+k. Find the value(s) of k for which f(x) has no real roots.
Solution-0.12<k<2.12 - f(x)=hx^2+kx-4h, h>0. Find the number of roots for the equation f(x)=0. Justify your answer.
Solution2 solutions because D>0 - f(x)=2(x+2)(x-4). y=kx-16 is tangent to this parabola. Find k.
Solution-4 - f(x)=ax^2-4x-c. A horizontal line, L, intersects the graph of f at x=-1 and x=3.
- Find the axis of symmetry.
Solutionx=1 - Find a.
Solutiona=2 - The equation of L is y=5. Find the value of c.
Solutionc=-1
- Find the axis of symmetry.
- f(x)=ax^2-12x+c
- A horizontal line, L, intersects f at x=2 and x=4
- Find the equation of the line of symmetry
Solutionx=3 - Hence show that a=2
- Find the equation of the line of symmetry
- The equation of L is y=-2. Find c.
Solution14
- A horizontal line, L, intersects f at x=2 and x=4
- See y=ax^2+bx+c below:
Positive, negative, or zero?- a
SolutionNegative - b
SolutionNegative - c
Solution0 - b^2-4ac
Solution0
- a
- \text{Revenue}=\text{num items}\times \text{price per item}. You normally sell 200 items at $50 each. For each $1 price increase you lose 2 sales.
- Define x given R=(200-2x)(50+x)
Solution# of price increases - Maximum revenue?
Solution\$11250 - What price should you sell each item to maximize revenue?
Solution\$75
- Define x given R=(200-2x)(50+x)
- The perimeter of the diagram below is 40.
- Show that y=20-\frac{\pi x}{2}
- Show that the area A=20x-\frac{5\pi x^2}{16}
- f(x)=(x-2)^2, x\geq2. Find f^{-1}(x)
- f(x)=x^2+2x-3, x<-1.
- Find f^{-1}{x}
- Evaluate f(-3)
- Evaluate f^{-1}(0)
- f(x)=x-3 and g(x)=x^2.
- Find f(g(x))
- Find (g\circ f)(x)
- Find (f\circ g)^{-1}(x)