What is the equation of the axis of symmetry for this function? H(x) = -X2 + 10x - 11 x = -5 0 x= -10 x = 5 x = 10

Mathematics

1 answer:

siniylev [52]3 years ago

8 0

Given:

The equation of a quadratic function is:

$H(x)=-x^2+10x-11$

To find:

The equation of the axis of symmetry for the given function.

Solution:

If a quadratic function is $f(x)=ax^2+bx+c$ , then the equation of the axis of symmetry is:

$x=-\dfrac{b}{2a}$ ...(i)

We have,

$H(x)=-x^2+10x-11$

Here, $a=-1,b=10,c=-11$ . So, the equation of axis of symmetry by using (i), is

$x=-\dfrac{10}{2(-1)}$

$x=-\dfrac{10}{-2}$

$x=5$

Therefore, the correct option is C.

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Write the equation of the line given points (-4,6) and (-8,10)

Advocard [28]

Answer:

the equation of the line in slope-intercept form is: $y=-x+2$

Step-by-step explanation:

First start by finding the slope of the segment that joins those two points using the general formula for the slope of the segments between two points $(x_1,y_1)$ and $(x_2,y_2)$ on the plane:

$Slope=\frac{y_2-y_1}{x_2-x_1}$

Then for our case, calling (-4, 6) = $(x_1,y_1)$ and (-8, 10) = $(x_2,y_2)$ , we have:

$Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{10-6}{-8-(-4)}\\Slope=\frac{4}{-8+4}\\Slope=\frac{4}{-4}\\Slope= -1$

Now, knowing this, we can find the equation of the line by using the "point-slope" form of a line [of slope "m" and going through the point $(x_0,y_0)$ that tells us: