Let $f(x)=ax^2+bx+c,~a\neq0$ . Its vertex can be found on the coordinates $(x_v,~y_v)$ such that $x_v=-\dfrac{b}{2a}$ and $y_v=-\dfrac{b^2}{4a}+c$ , in which $y_v=f(x_v)$ .

Using the coefficients given by the question, we get that:

$x_v=-\dfrac{8}{2\cdot(-1)}\\\\\\ x_v=-\dfrac{8}{-2}\\\\\\ x_v=4$

Thus, we have:

$y=f(x_v)=f(4)$

So the statement is true, because the $x$ coordinate of the vertex of the function is equal to $4.~~\checkmark$

7 0

3 years ago

Write a equation that represents the line

Phoenix [80]

Where’s the picture?

6 0

3 years ago

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Which is the simplified form of

AnnZ [28]

Answer: $\frac{1}{r^{7} } +\frac{1}{s^{12} }$

Step-by-step explanation:

With negative powers, $r^{-7} = \frac{1}{r^{7} }$

do the same for s too

6 0

3 years ago

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Write the equation of this circle in standard form. A) (x - 2)2 + (y - 3)2 = 4

Vika [28.1K]

Answer:

$x^2 + y^2 -4x -6y + 9 = 0$

Step-by-step explanation:

To write a circle in standard form, expand the equation of the circle and move all terms to one side.

$(x-2)^2 + (y-3)^2 = 4\\x^2 -4x + 4 + y^2 -6y + 9 = 4\\x^2 +y^2 - 4x - 6y + 13 = 4\\x^2 + y^2 -4x -6y + 9 = 0$

8 0

3 years ago