how would i do this, graphs http://prntscr.com/3usxns
@kirbykirby
@amoodarya
@kropot72
@nincompoop
Recall that you can write the equation of the parabola in the form \( a(x-h)^2+k\), where \(h\) represents the horizontal shifts and \(k\) represents the vertical shifts. The \(a\) is just the dilation factor, and your initial parabola is \(x^2\), so \(a=1\), and your parabola didn't undergo a dilation.. it just moved In fact, the point \((h, k)\) will the the coordinates of your vertex. So, can you determine \(h\) and \(k \)?
4 and 2
not quite exactly. You are in the quadrant where the x-values are negatives, and the y-values are negative as well
-4 and -2
yes, so now you can plug it in the formula \(a(x-h)^2+k\)
what is k
oh never mind
1(x-(-(-4))^2+(-2)
would it be 14
looks almost right but there is an extra minus (x - (-4))^2 + (-2)
oh it won't evaluate to a number.. this is the equation for the parabola. The x remains a variable with no fixed value
\[x-(-4)^2+(-2)\]
Be careful for the parenthesis, it is very important to keep it where it is \((x-(-4))^2 + (-2) \) is not the same as \(x-(-4)^2+(-2)\) It's probably best though to start simplifying all the negatives signs though to write \((x+4)^2-2\)
If you have an equation y = f(x), - when you replace x with x - h, the graph shifts h units horizontally (to the right if h is positive, and to the left if h is negative. - when you replace y with y - k, the graph shifts k units vertically (up if k is positive, and down if k is negative.
The blue graph is \(f(x) = x^2\) The red graph has shifted 4 units left and 2 units down. That means that x was replaced with x - (-4) = x + 4. Also, y is replaced with y - (-2) = y + 2 Rewriting f(x) as y, we get y = x^2 Applying the shifts, we get: \(y + 2 = (x + 4)^2\) Solving for y, we get: \(y = (x + 4)^2 - 2\) Now we replace y with f(x) to get: \(f(x) = (x + 4)^2 - 2\)
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