Notes: A curve has equation y= x^2 + 5x + 7(i) Find the coordinates of the vertex of the curve (ii) State the equation of the line of symmetry of the...

Saturday, 14 January 2017

A curve has equation y= x^2 + 5x + 7(i) Find the coordinates of the vertex of the curve (ii) State the equation of the line of symmetry of the...

Given $\displaystyle{{x}}^{{2}}+{5}{x}+{7}$ :

(i) Find the vertex:

(a) Rewrite in vertex form:

$\displaystyle{{x}}^{{2}}+{5}{x}+{7}$

Add and subtract the square of 1/2 the linear term, $\displaystyle{{\left(\frac{{5}}{{2}}\right)}}^{{2}}$ , to get $\displaystyle={{x}}^{{2}}+{5}{x}+\frac{{25}}{{4}}-\frac{{25}}{{4}}+{7}$

$\displaystyle={{\left({x}+\frac{{5}}{{2}}\right)}}^{{2}}+\frac{{3}}{{4}}$

is now in vertex form with vertex $\displaystyle{\left(-\frac{{5}}{{2}},\frac{{3}}{{4}}\right)}$

(b) Or find the axis of symmetry $\displaystyle{x}=\frac{{-{b}}}{{{2}{a}}}$ so $\displaystyle{x}=\frac{{-{5}}}{{{2}}}$ ; The value of the expression at this point is 3/4 so the vertex is at $\displaystyle{\left(-\frac{{5}}{{2}},\frac{{3}}{{4}}\right)}$

(ii) The...

Given $\displaystyle{{x}}^{{2}}+{5}{x}+{7}$ :

(i) Find the vertex:

(a) Rewrite in vertex form:

$\displaystyle{{x}}^{{2}}+{5}{x}+{7}$

Add and subtract the square of 1/2 the linear term, $\displaystyle{{\left(\frac{{5}}{{2}}\right)}}^{{2}}$ , to get $\displaystyle={{x}}^{{2}}+{5}{x}+\frac{{25}}{{4}}-\frac{{25}}{{4}}+{7}$

$\displaystyle={{\left({x}+\frac{{5}}{{2}}\right)}}^{{2}}+\frac{{3}}{{4}}$

is now in vertex form with vertex $\displaystyle{\left(-\frac{{5}}{{2}},\frac{{3}}{{4}}\right)}$

(ii) The line of symmetry is given by $\displaystyle{x}=\frac{{-{b}}}{{{2}{a}}}$ where the expression is given as $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}$ ; here a=1, b=5, and c=7. So the axis (line) of symmetry is $\displaystyle{x}=-\frac{{5}}{{2}}$ .

(iii) The graph is a parabola, opening up, normal width, with vertex $\displaystyle{\left(-\frac{{5}}{{2}},\frac{{3}}{{4}}\right)}$

The y-intercept can be found by setting x=0 to get 7. Some points (found by choosing convenient x-values near the vertex and symmetry) include (0,7),(-1,3),(-2,1),(-3,1), (-4,3),(-5,7)

The graph:

** The vertex form is $\displaystyle{y}={{\left({x}+\frac{{5}}{{2}}\right)}}^{{2}}+\frac{{3}}{{4}}$ . The 5/2 shifts the graph of $\displaystyle{y}={{x}}^{{2}}$ left 2.5 units; the $\displaystyle\frac{{3}}{{4}}$ term shifts the graph of $\displaystyle{y}={{x}}^{{2}}$ up $\displaystyle\frac{{3}}{{4}}$ of a unit. Since the leading coefficient is 1 there is no vertical stretch/compression.**