PLEASE SOMEONE HELP ME. Use the given values to write a quadratic equation of the form f(x)=ax^2+bx+c:

1 answer:

5 0

$\bf ~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \stackrel{vertex}{(2,-3)}~~ \begin{cases} h=2\\ k=-3 \end{cases}\implies y = a(x-2)^2-3 \\\\\\ \stackrel{passes~through}{(4,-1)}~~ \begin{cases} x=4\\ y=-1 \end{cases}\implies -1=a(4-2)^2-3\implies 2=a(4-2)^2 \\\\\\ 2=a(2)^2\implies 2=4a\implies \cfrac{2}{4}=a\implies \cfrac{1}{2}=a \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=\cfrac{1}{2}(x-2)^2-3~\hfill$

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Complete the square to rewrite y-x^2-6x+14 in vertex form. then state whether the vertex is a maximum or minimum and give its co

ycow [4]

Answer:

$y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2$

And solving we have:

$y= x^2 -6x +9 + 14 -9$

$y= (x-3)^2 +5$

And we can write the expression like this:

$y-5 = (x-3)^2$

The vertex for this case would be:

$V= (3,5)$

And the minimum for the function would be 3 and there is no maximum value for the function

Step-by-step explanation:

For this case we have the following equation given:

$y= x^2 -6x +14$

We can complete the square like this:

$y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2$

And solving we have:

$y= x^2 -6x +9 + 14 -9$

$y= (x-3)^2 +5$

And we can write the expression like this:

$y-5 = (x-3)^2$

The vertex for this case would be:

$V= (3,5)$

And the minimum for the function would be 3 and there is no maximum value for the function

4 0

3 years ago

How to find the vertex of a parabola given a quadratic equation?

Nat2105 [25]

$let \: f(x) = ax {}^{2} + bx + c \\ thus \: f(x) \: represents \: a \: parabola \: whose \: axis \: is \: parallel \: to \: the \: y \: axis \: and \: vertex \: is \: \frac{ - b}{2a} ,\frac{ - d}{4a} \\$

At the respective max and min values