Need help will medal topic is solving quadratic functions
Last one #22
you are given the v0 and s0 plug those in
\[h(t)=-16t^2+4t+10\] to find the max, find the vertex
@freckles ok one second let me find the vertex
@freckles ok I need help I can't figure it out
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@freckles will you help?
\[y=ax^2+bx+c \\ \text{ we are going to put this in the form } \\ y=a(x-h)^2+k \\ \] \[ \text{ So to makes things easier for you to see I'm going to multiply second term by } \frac{a}{a} \\ y=ax^2+bx+c \\ y=ax^2+\frac{a}{a}bx+c \\ \text{ Now from the first two terms I'm going to factor out a } a \\ y=a(x^2+\frac{1}{a}bx)+c \\ \text{ Now for completing the square } \\ y=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a (\frac{b}{2a})^2 \\ \text{ Notice here I added in a zero to complete the square } \\ \text{ that zero being } a (\frac{b}{2a})^2-a(\frac{b}{2a})^2 \\ \text{ now we can write this think as } \\ y=a(x+\frac{b}{2a})^2+c-a(\frac{b}{2a})^2\] So this in vertex form and the vertex is : \[(-\frac{b}{2a},c-a(\frac{b}{2a})^2)\]
what people usually this is calculate the x-coordinate of the vertex which is -b/(2a) then they plug into the original function so they don't have to remember the extra there... \[(-\frac{b}{2a},f(-\frac{b}{2a}))\]
where \[f(x)=ax^2+bx+c\]
So anyways see if you can find the x-coordinate now.
(1/8, 41/4) @freckles
yes that is what I have gotten
now the 1/8 says when the max happens 41/4 says what the max is
and I guess you should say 41/4 ft (including units make you awesome and more correct)
So let restate what I said (with the units) now 1/8 secs is when the max happens 41/4 ft is what the max is
Wait what? @freckles
the vertex is the maximum point the t-coordinate is where the max occurs the h-coordinate is what the max is I was just saying you may want to include units also
t-coordinate is in seconds units h-coordinate is in feet units
Okay so what would I put for my answer ? @freckles
hint: the question is asking for the maximum height
and I actually already said what this was above
1/8t or X=1/8? @freckles
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