Worked examples · Sheet 2

Parabolas with vertex, intercepts and graph

Exercise 1

Study the parabola

y = x^{2} - 4 x + 3

and draw its graph.

Concavity. $a = 1 > 0$ , therefore the parabola opens upward.

Vertex and axis.

x_{V} = - \frac{b}{2 a} = \frac{4}{2} = 2, y_{V} = 2^{2} - 4 \cdot 2 + 3 = - 1,

therefore $V (2, - 1)$ and the axis of symmetry is $x = 2$ .

Intersections with the axes. With the $x$ -axis:

x^{2} - 4 x + 3 = 0, x = \frac{4 \pm 16 - 12}{2} = \frac{4 \pm 2}{2},

therefore the points $(1, 0)$ and $(3, 0)$ . With the $y$ -axis: the point $(0, 3)$ .

y = x^{2} - 4 x + 3

y = x² − 4x + 3 with vertex V(2, −1)

Result: vertex $V (2, - 1)$ , axis $x = 2$ , intersections $(1, 0)$ , $(3, 0)$ and $(0, 3)$ .

The sign of the coefficient of $x^{2}$ is read first, because it determines the shape of the graph: here $a$ is positive, therefore the vertex is the lowest point. The abscissa of the vertex comes from the formula $x_{V} = - \frac{b}{2 a}$ , and the ordinate is obtained by substituting $x_{V}$ into the equation of the parabola, with no need for other formulas. The intersections with the $x$ -axis are the solutions of the associated equation, here two because the discriminant is positive; the intersection with the $y$ -axis is read from the constant term. With vertex, concavity and three points the parabola is determined and can be drawn.

Exercise 2

Study the parabola

y = - 2 x^{2} + 4 x + 6

and draw its graph.

Concavity. $a = - 2 < 0$ , therefore the parabola opens downward.

Vertex and axis.

x_{V} = - \frac{b}{2 a} = - \frac{4}{- 4} = 1, y_{V} = - 2 + 4 + 6 = 8,

therefore $V (1, 8)$ and the axis of symmetry is $x = 1$ .

Intersections with the axes. With the $x$ -axis, dividing by $- 2$ :

x^{2} - 2 x - 3 = 0, x = \frac{2 \pm 4 + 12}{2} = \frac{2 \pm 4}{2},

therefore the points $(- 1, 0)$ and $(3, 0)$ . With the $y$ -axis: the point $(0, 6)$ .

y = -2 x^{2} + 4 x + 6

y = −2x² + 4x + 6 with vertex V(1, 8)

Result: vertex $V (1, 8)$ , axis $x = 1$ , intersections $(- 1, 0)$ , $(3, 0)$ and $(0, 6)$ .

The coefficient $a = - 2$ is negative, therefore the parabola opens downward and the vertex is the highest point. For the intersections with the $x$ -axis it is convenient to divide the equation $- 2 x^{2} + 4 x + 6 = 0$ by $- 2$ , since the solutions do not change and the computations are done with smaller coefficients. The vertex has abscissa equidistant from the two roots, and indeed $1$ is the average of $- 1$ and $3$ ; this observation also provides a check on the computation of $x_{V}$ .

Exercise 3

Study the parabola

y = x^{2} + 2 x + 3

and draw its graph.

Concavity. $a = 1 > 0$ , therefore the parabola opens upward.

Vertex and axis.

x_{V} = - \frac{b}{2 a} = - 1, y_{V} = 1 - 2 + 3 = 2,

therefore $V (- 1, 2)$ and the axis of symmetry is $x = - 1$ .

Intersections with the axes. With the $x$ -axis: the discriminant is

Δ = 4 - 12 = - 8 < 0,

therefore the parabola does not intersect the $x$ -axis. With the $y$ -axis: the point $(0, 3)$ , and by symmetry with respect to the axis the point $(- 2, 3)$ also belongs to the graph.

y = x^{2} + 2 x + 3

y = x² + 2x + 3 with vertex V(−1, 2)

Result: vertex $V (- 1, 2)$ , axis $x = - 1$ , no intersection with the $x$ -axis, intersection $(0, 3)$ with the $y$ -axis.

The negative discriminant says that the equation $x^{2} + 2 x + 3 = 0$ has no real solutions, therefore the parabola does not touch the $x$ -axis; since it opens upward and the vertex has ordinate $2 > 0$ , the whole graph lies above the axis. To draw it, however, points other than the vertex are needed: the intersection with the $y$ -axis is $(0, 3)$ , and its symmetric point with respect to the line $x = - 1$ is $(- 2, 3)$ , because the two points have the same distance from the axis of symmetry.