Questions about y = ax²+bx+c.

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May 11th 2010, 05:55 PM
Divina

Questions about y = ax²+bx+c.

f(x) : ax²+bx+c, a≠ 0

[have to find the a(x-h)²+k form first]

Find an expression in terms of a, b, c for: (without using graph or calculator)
(i) An equation of the axis of symmetry
(ii) The maximum or minimum value
(iii) The coordinated of the vertex
(iv) The domain and the range
(v) The y-intercept of the graph of the function
(vi) The zeroes of the function

Discuss how you can predict the number of zeroes for a given quadratic function of the form y= ax²+bx+c, a≠ 0. Support the validity of your prediction with some examples.

My attempt
(ax² + bx + b/2 - b/2) + c
(ax² + bx + b/2) + c - b/2

(v) y intercept is (0,c)
May 11th 2010, 07:07 PM
pencil09

Quote:

Originally Posted by Divina

f(x) : ax²+bx+c, a≠ 0

[have to find the a(x-h)²+k form first]

find $a(x-h)^2+k$ form first....

$ax^2+b x+c=0$ (divided by $a$ )
$x^2+\frac{b}{a} x+ \frac {c}{a}=0$ (substract by $\frac {c}{a}$ )

$x^2+\frac{b}{a} x =-\frac {c}{a}$ (add by $(\frac {b}{2a})^2$ )

$(x+\frac{b}{2a})^2=-\frac {c}{a}+(\frac {b}{2a})^2$

$(x+\frac{b}{2a})^2+(\frac {c}{a}-(\frac {b}{2a})^2)=0$

so,
$h=-\frac{b}{2a}$

$k=(\frac {c}{a}-(\frac {b}{2a})^2)$
or $k=-\frac{b^2-4ac}{4a^2}$

-----I hope it'll help U-----
May 12th 2010, 01:29 AM
Divina

Quote:

Originally Posted by pencil09

find $a(x-h)^2+k$ form first....

$ax^2+b x+c=0$ (divided by $a$ )
$x^2+\frac{b}{a} x+ \frac {c}{a}=0$ (substract by $\frac {c}{a}$ )

$x^2+\frac{b}{a} x =-\frac {c}{a}$ (add by $(\frac {b}{2a})^2$ )

$(x+\frac{b}{2a})^2=-\frac {c}{a}+(\frac {b}{2a})^2$

$(x+\frac{b}{2a})^2+(\frac {c}{a}-(\frac {b}{2a})^2)=0$

so,
$h=-\frac{b}{2a}$

$k=(\frac {c}{a}-(\frac {b}{2a})^2)$
or $k=-\frac{b^2-4ac}{4a^2}$

-----I hope it'll help U-----

Thank you very much, that helped a lot