Coordinate Plane - ACT Math

Card 0 of 2502

Question

Solve the following system of equations:

–2x + 3y = 10

2x + 5y = 6

Answer

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)

Compare your answer with the correct one above

Question

What is the $y\textup{ intercept}$ of the following equation: ?

Answer

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

Compare your answer with the correct one above

Question

What is the equation for a circle of radius 12, centered at the intersection of the two lines:

y_1 = 4_x + 3

and

y_2 = 5_x + 44?

Answer

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

Compare your answer with the correct one above

Question

What is the equation for a circle of radius 9, centered at the intersection of the following two lines?

Answer

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 :

$y = 2 \cdot (-6) + 14$

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:

Compare your answer with the correct one above

Question

The diameter of a circle has endpoints at points (2, 10) and (–8, –14). Which of the following points does NOT lie on the circle?

Answer

Compare your answer with the correct one above

Question

The Y axis is a _______________ of the function Y = 1/X

Answer

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.

Compare your answer with the correct one above

Question

What is the domain of the following function:

$y=\frac{2-x}{x^2+5x+6}$

Answer

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

Compare your answer with the correct one above

Question

Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?

Answer

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

Compare your answer with the correct one above

Question

Give the equation, in slope-intercept form, of the line tangent to the circle of the equation

$x^{2}-6x + y^{2} - 91 = 0$

at the point .

Answer

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

$x^{2}-6x + y^{2} - 91 = 0$

$x^{2}-6x + y^{2} = 91$

Complete the square:

$x^{2}-6x+9 + y^{2} = 91+9$

$\left (x-3 \right )^{2} + y^{2} = 100$

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

$\frac{ y_{2} - y_{1} }{x_{2} - x_{1} } = \frac{8 -0}{-3-3} = \frac{8}{-6} = - \frac{4}{3}$ ,

so the tangent line has the opposite of the reciprocal of this, or $\frac{3}{4}$ , as its slope.

The tangent line therefore has equation

$y - 8= \frac{3}{4} (x- (-3))$

$y - 8= \frac{3}{4} (x+3)$

$y - 8= \frac{3}{4} x+\frac{9}{4}$

$y = \frac{3}{4} x+\frac{41}{4}$

Compare your answer with the correct one above

Question

Give the equation, in slope-intercept form, of the line tangent to the circle of the equation

$x^{2}+ y^{2} = 225$

at the point .

Answer

The graph of the equation $x^{2}+ y^{2} = 225$ 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

$\frac{ y_{2} - y_{1} }{x_{2} - x_{1} } =\frac{-9 -0}{12-0} =\frac{-9 }{12 }=- \frac{3}{4}$ ,

so the tangent line has the opposite of the reciprocal of this, or $\frac{4}{3}$ , as its slope.

The tangent line therefore has equation

$y - y_{1} = m (x-x_{1} )$

$y - (-9)= \frac{4}{3} (x-12 )$

$y +9= \frac{4}{3} x- 16$

$y = \frac{4}{3} x- 25$

Compare your answer with the correct one above

Question

What is the equation of a tangent line to

at point ?

Answer

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 $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find our slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}+4$

.

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug the point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation becomes,

Once we rearrange, the equation is

Compare your answer with the correct one above

Question

What is the equation of a tangent line to

at point

?

Answer

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 $y=ax^n \rightarrow y'=nax^{n-1}$ .

First we need to find our slope by plugging in our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=3x_{1}^2$

.

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ , we plug the point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Compare your answer with the correct one above

Question

Find the tangent line equation to

at point

?

Answer

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 $y=ax^n \rightarrow y'=nax^{n-1}$ .

First we need to find our slope at our $x_{1}$ by plugging in the value into our derivative equation and solving.

Thus, the slope is

$m=2x_{1}$

.

To find the equation of a tangent line of a given point $(x_{1},y_{1})$

We plug into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Compare your answer with the correct one above

Question

What is the tangent line equation of

at point

?

Answer

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 $y=ax^n \rightarrow y'=nax^{n-1}$ .

First we need to find the slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=4x_{1}^3+4$

.

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug it into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Compare your answer with the correct one above

Question

Find the equation of a tangent line to

for the point

?

Answer

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 $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find the slope by plugging in our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}$

.

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug our point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Compare your answer with the correct one above

Question

What is the equation of a tangent line to

at the point

?

Answer

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 $y=ax^n \rightarrow y'=nax^{n-1}$ .

First we need to find our slope by plugging in our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=25x_{1}^4$

.

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug the point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Compare your answer with the correct one above

Question

Find the equation of a tangent line to

at the point

?

Answer

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 $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find the slope by plugging in our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}$

.

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug our point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Compare your answer with the correct one above

Question

What is the equation of a tangent line to

at the point

?

Answer

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 $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find our slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}+1$

.

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug our point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Compare your answer with the correct one above

Question

What is the equation of a tangent line to

at the point

?

Answer

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 $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find our slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}+5x_{1}^4$

.

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug our point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Compare your answer with the correct one above

Question

What is the equation of a tangent line to

at the point

?

Answer

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 $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find the slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}$

.

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug our point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Compare your answer with the correct one above

Tap the card to reveal the answer