Complete the square to transform the quadratic equation into the form (x – p)2 = q.

1 answer:

3 0

If this problem is expecting you to show a graph photo, then a dependable website for a problem like this would be www.desmos.com. message me if you're not sure how to use it.

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A baseball is thrown straight upwards from the ground and undergoes a free fall motion as it rises towards its highest point. Wh

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Answer: The answer should be that the velocity and acceleration decreases ! :)

Step-by-step explanation: With velocity, the pure definitions are "quickness of motion," and "rapidity of movement." So with this, when you think about it, the speed/velocity of the baseball decreases the higher it goes because of gravity slowly getting stronger the higher up it goes. So the answer would be that the velocity and the acceleration decreases. (hope I helped :') )

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What is equivalent to the expression?<img src="https://tex.z-dn.net/?f=4x%5E%7B2%7D%20%2A4x%5E%7Bx%7D" id="TexFormula1" title="4

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Answer:

Step-by-step explanation:

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2 years ago

The graph of f(x) = x^2 has been shifted into the form f(x) = (x − h)^2 + k

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind, let's see, it went to the right 2 units, and then up 3 units.

that simply means, C = -2, D = 3.

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The answer is 825

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