2.3 Quadratic & Rational Functions

The diagram shows part of the graph of `y=a(x-h)^2+k`. The graph has its vertex at `P`, and passes through the point `A` with coordinates `(1,0)`.

(a) Write down the value of

    (i) `h`;

    (ii) `k`.

(b) Calculate the value of `a`.

(a) (i) `h=-1`

     (ii) `k=2`

(b) `a(1+1)^2+2=0`

     `a=-0.5`

Consider the following function:

                   `h(x)=2/(sqrt(x-1))+1/2," for " x>1.`

(a) Find `h^(-1) (1)`

(b) Find the domain of `h^(-1) (x)`

(a) setting `h(x)=1`

     `h^(-1) (1)=17`

(b) `x>1/2`

The diagram shows part of the graph of the curve `y=a(x-h)^2+k`, where `a,h,k∈Z`

(a) The vertex is at the point `(3,1)`. Write down the value of `h` and of `k`.

(b) The point `P(5,9)` is on the graph. Show that `a=2`.

(c) Hence show that the equation of the curve can be written as `y=2x^2-12x+19.`

(d) (i) Find `dy/dx`

     A tangent is drawn to the curve at `P(5,9)`

     (ii) Calculate the gradient of this tangent.

     (iii) Find the equation of this tangent.

(a) Since the vertex is at `(3,1)`

      `h=3`

      `k=1`

(b) `(5,9)` is on the graph

     `⇒9=a(5-3)^2+1`

     `=4a+1`

     `⇒9-1=4a=8`

     `⇒a=2`

(c) `y=2(x-3)^2+1`

      `=2x^2-12x+19`

(d) (i) Graph has equation `y=2x^2-12x+19`

                                        `dy/dx=4x-12` 

     (ii) At point `(5,9)`, gradient `=4(5)-12=8`

     (iii) Equation: `y-9=8(x-5)`

                          `8x-y-31=0`

(a) The diagram shows part of the graph of a quadratic function `f(x)=x^2+bx+c`, which intersects the x-axis at `x=2` and at `x=3`

      Find the value of `b` and of `c`

(b) The diagram shows part of the graph of another quadratic function `g`. It can be written in the form `g(x)=a(x-h)^2+3`. Its vertex is at `(2,3)` and its y-intercept is `5`.

     (i) Write down the value of `h` 

     (ii) Find the value of `a`

(a)      `b=-5, c=6`

(b) (i) `h=2`

     (ii) `g(x)=a(x-2)^2+3`

          `5=a(0-2)^2+3`

          `a=0.5`

The following diagram shows part of the graph of `f`, where `f(x)=x^2-x-2`

(a) Find both x-intercepts.

(b) Find the x-coordinate of the vertex.

(a) evidence of attempting to solve `f(x)=0`

     evidence of correct working

     e.g. `(x+1)(x-2), (1±sqrt9)/2`

     intercepts are `(-1,0)` and `(2,0)` (accept `x=-1, x=2`)

(b) evidence of appropriate method

     e.g. `x_v=(x_1+x_2)/2, x_v=(-b)/2a`, reference to symmetry

     `x_v=0.5`

A function `f` is defined by `f(x)=(2x-1)/(x+1)`, where `x∈R,x≠-1`

(a) The graph of `y=f(x)` has a vertical asymptote and a horizontal asymptote.

      Write down the equation of

     (i) The vertical asymptote;

     (ii) The horizontal asymptote.

(b) On the set of axes below, sketch the graph of `y=f(x)`

     On your sketch, clearly indicate the asymptotes and the position of any points of intersection with the axes.

(c) Hence, solve the inequality `0<(2x-1)/(x+1)<2`

(a) (i) `x=-1`

     (ii) `y=2`

     axes intercepts clearly shown at `x=1/2` and `y=-1`

(b)

(c) `x>1/2`

Let `f(x)=p(x-q)(x-r)`. Part of the graph of `f` is shown below.

The graph passes through the points `(-2,0)`, `(0,-4)` and `(4,0)`

(a) Write down the value of `q` and of `r`

(b) Write down the equation of the axis of symmetry.

(c) Find the value of `p`

(a) `q=-2, r=4` or `q=4, r=-2`

(b) `x=1` (must be an equation)

(c) substituting `(0,-4)` into the equation

     e.g. `-4=p(0-(-2))(0-4)`, `-4=p(-4)(2)`

     correct working towards solution

     e.g.  `-4=-8p`

     `p=4/8 (=1/2)`

 

Let `f(x)=1/2 x^2+kx+8`, where `k∈Z`

(a) Find the values of `k` such that `f(x)=0` has two equal roots.

(b) Each value of `k` is equally likely for `-5≤k≤5`. Find the probability that `f(x)=0` has no roots.

(a) Recognising that equal roots means perfect square

     e.g. attempt to complete the square, `1/2(x^2+2kx+16)`

     correct working

     e.g. `1/2(x+k)^2, 1/2 k^2=8`

     `k=±4`

(b) evidence of appropriate approach

     e.g. `b^2-4ac<0`

     correct working for `k`

     e.g. `-4ltklt4`, `k^2 lt6`

     `p=7/11`

Let `f` be a quadratic function. Part of the graph of `f` is shown below.

The vertex is at `P(4,2)` and the `y`-intercept is at `Q(0,6)`

(a) Write down the equation of the axis of symmetry.

The function `f` can be written in the form `f(x)=a(x-h)^2+k`

(b) Write down the value of `h` and  of `k`.

(c) Find `a`

(a) `x=4` (must be an equation)

(b) `h=4`, `k=2`

(c) attempt to substitute coordinates of any point on the graph into `f`

     e.g. `f(0)=6, 6=a(0-4)^2+2, f(4)=2`

     correct equation (do not accept an equation that results from `f(4)=2`

     e.g. `6=a(-4)^2+2, 6=16a+2` 

            `a=4/16 (=1/4)`

A function `f` is defined by `f(x)=(2(x+3))/(3(x+2))`, where `x∈R,x≠-2`

The graph `y=f(x)` is shown below.

(a) Write down the equation of the horizontal asymptote.

Consider `g(x)=mx+1`, where `m∈R,m≠0`

(b) (i) Write down the number of solutions to `f(x)=g(x)` for `m>0`

     (ii) Determine the value of m such that `f(x)=g(x)` has only one solution for `x` 

     (iii) Determine the range of values for `m`, where `f(x)=g(x)` has two solutions for `x≥0`

(a) `y=2/3` (must be written as equation with `y=`)

(b) (i) `2`

     (ii) `(2(x+3))/(3(x+2))=mx+1`

     attempt to expand to obtain a quadratic equation

     `2x+6=3mx^2+6mx+3x+6`

     `3mx^2+(6m+1)x=0 "OR" 3mx^2+6mx+x=0`

     attempt to solve their quadratic for `x` and equating their solutions

     `x(3mx+6m+1)=0`

     `x=0 "OR" x=(-6m+1)/3m (=0)`  

     `(-6m+1)/3m=0`

     `⇒m=-1/6`

     (iii) `-1/6ltmlt0`