Find the vertex of the parabola formed by the following quadratic function: y = 3x^2 – 6x + 2

1 answer:

7 0

Vertex form is y=a(x-h)^2+k, so we can rearrange to that form...

y=3x^2-6x+2 subtract 2 from both sides

y-2=3x^2-6x divide both sides by 3

(y-2)/3=x^2-2x, halve the linear coefficient, square it, add it to both sides...in this case: (-2/2)^2=1 so

(y-2)/3+1=x^2-2x+1 now the right side is a perfect square

(y-2+3)/3=(x-1)^2

(y+1)/3=(x-1)^2 multiply both sides by 3

y+1=3(x-1)^2 subtract 1 from both sides

y=3(x-1)^2-1 so the vertex is:

(1, -1)

...

Now if you'd like you can commit to memory the vertex point for any parabola so you don't have to do the calculations like what we did above. The vertex of any quadratic (parabola), ax^2+bx+c is:

x= -b/(2a), y= (4ac-b^2)/(4a)

Then you will always be able to do a quick calculation of the vertex :)

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What is the GCF for: 3x^3 -4x^2

serg [7]

3x^3 - 4x^2
x^2(3x - 4)

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Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g

SpyIntel [72]

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

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

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

with that template in mind, let's see $\bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad B
\end{array}$

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.

8 0

3 years ago

Read 2 more answers

What would be the answer please explain

sp2606 [1]

Using the line of the best fit, the predicted student's score in English test is 48

<h3>How to determine the student's score in English?</h3>

From the question, we have:

Mathematics score = 60

The scores in English tests are plotted on the y-axis.

On the given graph, we have:

(x,y) = (60,48)