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2 years ago

What are the coordinates of the vertex of the parabola with the equation y=x^2+2x-3

Mathematics

2 answers:

Gala2k [10]2 years ago

6 0

Answer: Answer is (-1,-4).

Studentka2010 [4]2 years ago

5 0

$\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+2}x\stackrel{\stackrel{c}{\downarrow }}{-3} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 2}{2(1)}~~~~ ,~~~~ -3-\cfrac{ (2)^2}{4(1)}\right)\implies \left( -1~~,~~-3-1 \right)\implies (-1~~,~~-4)$

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Answer:

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Step-by-step explanation:

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4 years ago

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4 years ago

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Write the quadratic function in the form g(x)=a (x-h)2 +kThen, give the vertex of its graph. G(x) =2x2 +20x+49Writing in the for

Snezhnost [94]

The quadratic function given to us is:

$g(x)=2x^2+20x+49$

We are asked to find the vertex form of the function.

The general formula for the vertex form of a quadratic equation is:

$\begin{gathered} g(x)=a(x-h)^2+k \\ \text{where,} \\ (h,k)\text{ is the coordinate of the vertex} \end{gathered}$

In order to write the function in its vertex form, we need to perform a couple of operations on the function.

1. Add and subtract the square of the half of the coefficient of x to the function.

2. Factor out the function with its repeated roots and re-write the equation.

Now, let us solve.

1. Add and subtract the square of the half of the coefficient of x to the function.

$\begin{gathered} g(x)=2x^2+20x+49=2(x^2+10x+\frac{49}{2}) \\ \text{half of coefficient of x:} \\ \frac{10}{2}=5 \\ \text{square of the half of the coefficient of x:} \\ 5^2=25 \\ \\ \therefore g(x)=2(x^2+10x+25-25+\frac{49}{2}) \end{gathered}$

2. Factor out the function with its repeated roots and re-write the equation.

$\begin{gathered} g(x)=2(x^2+10x+25)-2(25+\frac{49}{2}) \\ re-\text{write the above function} \\ g(x)=2(x^2+10x+25)-1 \\ \text{Let us factorize this:} \\ g(x)=2(x+5)^2-1 \end{gathered}$

Therefore, we can conclude that the Equation and vertex of the equation is:

$\begin{gathered} Equation\colon g(x)=2(x+5)^2-1 \\ \\ Vertex\colon(-5,-1) \end{gathered}$

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1 year ago

Determine the equation of the line that passes through the points of intersection of the graphs of the quadratic functions f(x)

timofeeve [1]

Answer: $x-2y-2=0$

Step-by-step explanation:

Given :

$f(x) = x^2 - 4 \\ g(x) = - 3x^2 + 2x + 8$

Point of intersection :

$f(x)=g(x)\\x^2-4=-3x^2+2x+8\\4x^2-2x-12=0\\2x^2-x-6=0\\2x^2-4x+3x-6=0\\2x(x-2)+3(x-2)=0\\(x-2)(2x+3)=0\\x=2\,,\,\frac{-3}{2}$

$x=2\,;f(2)=2^2-4=0\\x=\frac{-3}{2}\,; f\left ( \frac{3}{2} \right )=\left ( \frac{3}{2} \right )^2-4=\frac{-7}{4}=-1.75$

So, we have points $\left ( 2,0 \right )\,,\,\left ( -1.5,-1.75\ \right )$

Equation of line passing through two points $\left ( x_1,y_1 \right )\,,\,\left ( x_2,y_2 \right )$ is given by $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left ( x-x_1 \right )$