Coordinate Plane - ACT Math

This question relies on both the vertical-line test and the definition of a function. We need to use the vertical-line test to determine which of the graphs is not a function (i.e. the graph that has more than one output for a given input). The vertical-line test states that a graph represents a function when a vertical line can be drawn at every point in the graph and only intersect it at one point; thus, if a vertical line is drawn in a graph and it intersects that graph at more than one point, then the graph is not a function. The circle is the only answer choice that fails the vertical-line test, and so it is not a function.

There are four shifts of the graph y = f(x):

y = f(x) + c shifts the graph c units upwards.

y = f(x) – c shifts the graph c units downwards.

y = f(x + c) shifts the graph c units to the left.

y = f(x – c) shifts the graph c units to the right.

The graph has x-intercepts at x = 0 and x = 8. This indicates that 0 and 8 are roots of the function.

The function must take the form y = x(x - 8) in order for these roots to be true.

The parabola opens downward, indicating a negative leading coefficient. Expand the equation to get our answer.

y = -x(x - 8)

y = -x2 + 8x

y = 8x - x2

Therefore, the answer must be y = 8x - x2

One of the properties of taking an absolute value of a function is that the values are all made positive. The values themselves do not change; only their signs do. In this graph, none of the y-values are negative, so none of them would change. Thus the two graphs should be identical.

The basic formula for a circle in the coordinate plane is , where is the center of the circle with radius .

We know that , since that is the only way a diameter can pass through the circle and intercept an x-coordinate of at both ends. , on the other hand, may be seen as halfway between one y-coordinate and the other y-coordinate. Averaging the two, we get:

, so becomes our . Since the diameter is units long, we know the radius is half that, so .

Thus, we have .

Since we have –2x and +2x in the equations, it makes sense to add the equations together to give 8y = 16 yielding y = 2. Then we substitute y = 2 into one of the original equations to get x = –2. So the solution to the system of equations is (–2, 2)

The formula for a circle centered around a point with radius is given by:

.

Thus we see the answer is

The y-intercept is the constant at the end of the equation. Thus for our equation the y-intercept is 7

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

4_x_ + 3 = 5_x_ + 44; 3 = x + 44; –41 = x

To find the y-coordinate, substitute into one of the equations. Let's use _y_1:

y = 4 * –41 + 3 = –164 + 3 = –161

The center of our circle is therefore: (–41, –161).

Now, recall that the general form for a circle with center at (_x_0, _y_0) is:

(x - _x_0)2 + (y - _y_0)2 = _r_2

For our data, this means that our equation is:

(x + 41)2 + (y + 161)2 = 122 or (x + 41)2 + (y + 161)2 = 144

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

To find the y-coordinate, substitute into one of the equations. Let's use :

The center of our circle is therefore .

Now, recall that the general form for a circle with center at is

For our data, this means that our equation is:

The general formula for a circle centered about points with a radius of is:

.

Since we are centered about the origin both and are zero. Thus the equation we have is:

after simplifying

A line is an asymptote in a graph if the graph of the function nears the line as X or Y gets larger in absolute value.

The denominator cannot be zero, otherwise the function is indefinite. Therefore x cannot be –2 or –3.

The basic formula for a circle in the coordinate plane is , where is the center of the circle with radius .

Using this, we can simply substitute for , for , and for . Customarily, is simplified for the final equation.

----> .

The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by

The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.

The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.

The equation of the line is y – 4 = (3/4)(x – (–3))

Rearranging gives us: 3_x_ – 4_y_ = -25

Rewrite the equation of the circle in standard form to find its center:

Complete the square:

The center is .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints and will have slope

,

so the tangent line has the opposite of the reciprocal of this, or , as its slope.

The tangent line therefore has equation

The graph of the equation is a circle with center .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints and will have slope

,

so the tangent line has the opposite of the reciprocal of this, or , as its slope.

The tangent line therefore has equation

To find an equation tangent to

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

due to power rule .

First we need to find our slope by plugging our into the derivative equation and solving.

Thus, the slope is

.

To find the equation of a tangent line of a given point we plug the point into

.

Therefore our equation becomes,

Once we rearrange, the equation is

To find an equation tangent to

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

due to power rule .

First we need to find our slope by plugging in our into the derivative equation and solving.

Thus, the slope is

.

To find the equation of a tangent line of a given point , we plug the point into

.

Therefore our equation is

Once we rearrange, the equation is