Quadratic function of Y=x2-2x-3

1 answer:

6 0

Ok! So, given a quadratic function, y = ax2 + bx + c, when "a" is positive, the parabola opens upward and the vertex is the minimum value. On the other hand, if "a" is negative, the graph opens downward and the vertex is the maximum value. Now, let's refer back to our original graph, y = x2, where "a" is 1.

Hope this helps.

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Which matrix represents the system of equations shown below? 3x-5y=12 4x-2y=15

tatuchka [14]

Answer:

$\left[\begin{array}{ccc}3&-5 &|12\\4&-2 &|15\\\end{array}\right]$

Step-by-step explanation:

When making a matrix of two equations with the variables x and y, the result will be a matrix with three columns:

a column for the values of x in each equation
a column for the values of y in each equation
a column for the independent values of each equation

since our system of equations is:

$3x-5y=12\\ 4x-2y=15$

we can see that the value for x in the first equation is 3 and in the second equation is 4, thus the first column will have the numbers 3 and 4:

$\left[\begin{array}{ccc}3&&\\4&&\\\end{array}\right]$

Now for the values of y we hvae -5 in the first equation and -2 in the second equation, we update the matrix with another column with the values of -5 and -2:

$\left[\begin{array}{ccc}3&-5&\\4&-2&\\\end{array}\right]$

Finally, the last column is the independent values of each equation (or the results) in the first equation that number is 12 and in the second equation is 15, thus the matrix is:

$\left[\begin{array}{ccc}3&-5&12\\4&-2&15\\\end{array}\right]$