Find the vertex and the axis of symmetry of the graph of the function: f(x) = -2x2 – 4x?
The first step would be to find the vertex. Do you know how to write a given parabola in vertex form?
no
One way is to complete the square. It allows you to rewrite the standard form \(ax^2+bx+c\) into vertex form \((x-h)^2+k\), where \((h,k)\) is the vertex. So we're given \[y=-2x^2-4x\] Do you know how to complete the square?
is it -1 and-4?
aaa or -6 -4
\[\begin{align*}y&=-2x^2-4x\\ &=-2\left(x^2+2x\right)\\ &=-2\left(x^2+2x+1-1\right)\\ &=-2\left((x+1)^2-1\right)\\ &=-2(x+1)^2+2\\ \end{align*}\] What's the vertex?
lost because is not the answer in the back of my book
Well, according to my work, the vertex is \((-1,2)\), and Wolfram agrees: http://www.wolframalpha.com/input/?i=vertex+of+-2x%5E2-4x
sorry looks like i didnt put this at the end of the question What are the x-intercepts of y = f(x-7
of \(y=f(x-7)\) or \(f(x)-7\) ?
the first one
Okay. If \(f(x)=-2x^2-4x\), then what is \(f(x-7)\) ?
no clue lol
Plug in \(x-7\): \[\begin{align*}f(x)=-2x^4-4x~~\Rightarrow~~f(x-7)&=-2(x-7)^2-4(x-7)\\ &=-2\left(x^2-14x+49\right)-4x+28\\ &=-2x^2+28x-98-4x+28\\ &=-2x^2+24x-70\end{align*}\] Make sense so far?
yes thanks
the problem is thisFind the vertex and the axis of symmetry of the graph of the function: f(x) = -2x^2 – 4x
and this are the answers (-1,0); x = -1 (-1,2); x = -1 (1,2); x = 1 (1,0); x = 1
which am still not sure off
Hmm, something's not working with the equation editor. Give me a moment to write it out.
k thanks
|dw:1378269651664:dw| So what's the vertex?
1-1
Hmm? You should be getting a vertex of \((6,2)\)... http://www.wolframalpha.com/input/?i=vertex+of+-2%28x-7%29%5E2-4%28x-7%29
mmm
i think the book is confusing me
Answer key could be mismatched with a different question.
yea
thanks though
yw
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