Question

I have a question and hopefully someone can explain this... I have to do the following:

Create two additional quadratic functions, g(x) and h(x).
The function g(x) will open the same direction as f(x), have the same vertex, but will be narrower.
The function h(x) will open in the opposite direction as f(x), have the same vertex, but will be wider than f(x).

My equation is f(x)=4(x-1)^2+28

Answer 1

do you know what the vertex of \(\bf f(x)=4(x-1)^2+28\) is ?

Answer 2

would it be (-1, 28) ?

Answer 3

well... just (1, 28) yes, the " - " is part of the TEMPLATE, and so is the " + "

Answer 4

okay, so how to I begin then ?

Answer 5

\( \begin{array}{rllll}

&\downarrow &\downarrow \\
f(x)=\square (x-&\square )^2+&\square\\ \quad \\

g(x)=\bbox[border: 1px solid black]{{\color{red}{ \textit{bigger value}}}} (x-&\square )^2+&\square\qquad \textit{same vertex, narrower}\\ \quad \\

h(x)=\bbox[border: 1px solid black]{{\color{red}{ \textit{smaller value}}}} (x-&\square )^2+&\square \qquad \textit{same vertex, opens opposite, wider}
\end{array}\)

Answer 6

ahemm...

Answer 7

thank you so much !

Answer 8

\( \begin{array}{rllll}

&\downarrow &\downarrow \\
f(x)=\square (x-&\square )^2+&\square\\ \quad \\

g(x)=\bbox[border: 1px solid black]{{\color{red}{ \textit{bigger value}}}} (x-&\square )^2+&\square\\ \textit{same vertex, narrower}\\ \quad \\

h(x)=\bbox[border: 1px solid black]{{\color{red}{ \textit{smaller and negative value}}}} (x-&\square )^2+&\square \\ \textit{same vertex, opens opposite, wider}
\end{array}\)

Answer 9

if the value of the number in front of the binomial is smaller, say 1/2 or 1/4 or so, the parabola gets WIDER

if the value is bigger, the parabola gets NARROWER

if it's negative, the parabola gets flipped to the other side, in this case, upside-down