Mr Petros Tsichlakis

Sketching a quadratic graph

Sketch the graph of the following quadratic functionf(x) or y = -5+x-x2

Show clearly the coordinates of the points where the curve crosses both the coordinate axes, the line of symmetry and the coordinates of the vertex.

Mr Petros Tsichlakis

Model Answer

1) SHAPESince a = -1 is negative, our parabola is a ∩ shape.

2) y-crossingSince c = -5, our curve crosses the y axis at y=-5

Alternatively: We set x=0 in the equation y = -5+x-x2 and this will give us the value of y: f(0) = y = -5+0 – 02 = -5

Coordinates of y-crossing (0,-5)

Our equation is a quadratic of the form y=ax2+bx+c therefore a=-1 b=1 and c=-5

Mr Petros Tsichlakis

3) x-crossings ---We set y = 0 in the equation y = -5+x-x2 and this will give us the value(s) of x: 0 = -5+x – x2

This is a quadratic equation we need to solve. If we rearrange it we get:– x2 +x -5 = 0

Using the quadratic formula with a=-1, b=1 and c=-5 we find

Discriminant Δ= = (1)2-4*(-1)*(-5)=1-20=-19 which means we have no roots/solutions and our curve DOES NOT CROSS the x-axis!!

4) Since the VERTEX is needed, the best method we can use is COMPLETING THE SQUARE. This will reveal the coordinates of the vertex AND the line of symmetry!

By completing the square for – x2 +x -5 we find…..

Mr Petros Tsichlakis

which when compared to

gives a = -1 p = and q =

---Coordinates of VERTEX (max, since ∩ shape) :(-p, q) or ( , )opposite of p

---Line of symmetry : x = -p or x =

We are now ready to sketch the curve….

Mr Petros Tsichlakis

y

x0

---------------------

--------------

+12

−194 ( , )

line of symmetry x =

−5