### Key Concepts

**Distance Formula:**The distance*d*between the two points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ is

$$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$$**Midpoint Formula:**The midpoint of the line segment whose endpoints are the two points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ is

$$\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$$

To find the midpoint of a line segment, we find the average of the*x*-coordinates and the average of the*y*-coordinates of the endpoints.**Circle:**A circle is all points in a plane that are a fixed distance from a fixed point in the plane. The given point is called the*center,*$\left(h,k\right),$ and the fixed distance is called the*radius, r,*of the circle.**Standard Form of the Equation a Circle:**The standard form of the equation of a circle with center, $\left(h,k\right),$ and radius,*r,*is

**General Form of the Equation of a Circle:**The general form of the equation of a circle is

$${x}^{2}+{y}^{2}+ax+by+c=0$$

**Parabola:**A**parabola**is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the**focus,**and the fixed line is called the**directrix**of the parabola.

Vertical Parabolas General form

$y=a{x}^{2}+bx+c$Standard form

$y=a{\left(x-h\right)}^{2}+k$**Orientation**$a>0$ up; $a<0$ down $a>0$ up; $a<0$ down **Axis of symmetry**$x=-\frac{b}{2a}$ $x=h$ **Vertex**Substitute $x=-\frac{b}{2a}$ and

solve for*y*.$\left(h,k\right)$ *y*- interceptLet $x=0$ Let $x=0$ *x*-interceptsLet $y=0$ Let $y=0$

**How to graph vertical parabolas $(y=a{x}^{2}+bx+c$ or $f\left(x\right)=a{\left(x-h\right)}^{2}+k)$ using properties.**- Step 1. Determine whether the parabola opens upward or downward.
- Step 2. Find the axis of symmetry.
- Step 3. Find the vertex.
- Step 4.
Find the
*y*-intercept. Find the point symmetric to the*y*-intercept across the axis of symmetry. - Step 5.
Find the
*x*-intercepts. - Step 6. Graph the parabola.

Horizontal Parabolas General form

$x=a{y}^{2}+by+c$Standard form

$x=a{\left(y-k\right)}^{2}+h$**Orientation**$a>0$ right; $a<0$ left $a>0$ right; $a<0$ left **Axis of symmetry**$y=-\frac{b}{2a}$ $y=k$ **Vertex**Substitute $y=-\frac{b}{2a}$ and

solve for*x*.$\left(h,k\right)$ *y*-interceptsLet $x=0$ Let $x=0$ *x*-interceptLet $y=0$ Let $y=0$

**How to graph horizontal parabolas $(x=a{y}^{2}+by+c$ or $x=a{\left(y-k\right)}^{2}+h)$ using properties.**- Step 1. Determine whether the parabola opens to the left or to the right.
- Step 2. Find the axis of symmetry.
- Step 3. Find the vertex.
- Step 4.
Find the
*x*-intercept. Find the point symmetric to the*x*-intercept across the axis of symmetry. - Step 5.
Find the
*y*-intercepts. - Step 6. Graph the parabola.

**Ellipse:**An**ellipse**is all points in a plane where the sum of the distances from two fixed points is constant. Each of the fixed points is called a**focus**of the ellipse.

If we draw a line through the foci intersects the ellipse in two points—each is called a**vertex**of the ellipse.

The segment connecting the vertices is called the**major axis**.

The midpoint of the segment is called the**center**of the ellipse.

A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the**minor axis**.**Standard Form of the Equation an Ellipse with Center**$\left(0,0\right):$ The standard form of the equation of an ellipse with center $\left(0,0\right),$ is

$$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$$

The*x*-intercepts are $\left(a,0\right)$ and $\left(\text{\u2212}a,0\right).$

The*y*-intercepts are $\left(0,b\right)$ and $\left(0,\text{\u2212}b\right).$**How to an Ellipse with Center**$\left(0,0\right)$- Step 1. Write the equation in standard form.
- Step 2. Determine whether the major axis is horizontal or vertical.
- Step 3. Find the endpoints of the major axis.
- Step 4. Find the endpoints of the minor axis
- Step 5. Sketch the ellipse.

**Standard Form of the Equation an Ellipse with Center**$\left(h,k\right):$ The standard form of the equation of an ellipse with center $\left(h,k\right),$ is

$$\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$$

When $a>b,$ the major axis is horizontal so the distance from the center to the vertex is*a*.

When $b>a,$ the major axis is vertical so the distance from the center to the vertex is*b*.

**Hyperbola:**A**hyperbola**is all points in a plane where the difference of their distances from two fixed points is constant.

Each of the fixed points is called a**focus**of the hyperbola.

The line through the foci, is called the**transverse axis**.

The two points where the transverse axis intersects the hyperbola are each a**vertex**of the hyperbola.

The midpoint of the segment joining the foci is called the**center**of the hyperbola.

The line perpendicular to the transverse axis that passes through the center is called the**conjugate axis**.

Each piece of the graph is called a**branch**of the hyperbola.

Standard Forms of the Equation a Hyperbola with Center $\left(0,0\right)$ $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ $\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1$ **Orientation**Transverse axis on the *x*-axis.

Opens left and rightTransverse axis on the *y*-axis.

Opens up and down**Vertices**$\left(\text{\u2212}a,0\right),$ $\left(a,0\right)$ $\left(0,\text{\u2212}a\right),$ $\left(0,a\right)$ *x*-intercepts$\left(\text{\u2212}a,0\right),$ $\left(a,0\right)$ none *y*-interceptsnone $\left(0,\text{\u2212}a\right)$, $\left(0,a\right)$ **Rectangle**Use $\left(\text{\xb1}a,0\right)$ $\left(0,\text{\xb1}b\right)$ Use $\left(0,\text{\xb1}a\right)$ $\left(\text{\xb1}b,0\right)$ **asymptotes**$y=\frac{b}{a}x,$ $y=-\frac{b}{a}x$ $y=\frac{a}{b}x,$ $y=-\frac{a}{b}x$ **How to graph a hyperbola centered at $\left(0,0\right).$**- Step 1. Write the equation in standard form.
- Step 2. Determine whether the transverse axis is horizontal or vertical.
- Step 3. Find the vertices.
- Step 4. Sketch the rectangle centered at the origin intersecting one axis at $\text{\xb1}a$ and the other at $\text{\xb1}b.$
- Step 5. Sketch the asymptotes—the lines through the diagonals of the rectangle.
- Step 6. Draw the two branches of the hyperbola.

Standard Forms of the Equation a Hyperbola with Center $\left(h,k\right)$ $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ $\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1$ **Orientation**Transverse axis is horizontal.

Opens left and rightTransverse axis is vertical.

Opens up and down**Center**$\left(h,k\right)$ $\left(h,k\right)$ **Vertices***a*units to the left and right of the center*a*units above and below the center**Rectangle**Use *a*units left/right of center*b*units above/below the centerUse *a*units above/below the center*b*units left/right of center**How to graph a hyperbola centered at $\left(h,k\right).$**- Step 1. Write the equation in standard form.
- Step 2. Determine whether the transverse axis is horizontal or vertical.
- Step 3. Find the center and $a,b.$
- Step 4. Sketch the rectangle centered at $\left(h,k\right)$ using $a,b.$
- Step 5. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices.
- Step 6. Draw the two branches of the hyperbola.

Conic Characteristics of ${x}^{2}\text{-}$ and ${y}^{2}\text{-}$ terms Example **Parabola**Either ${x}^{2}$ OR ${y}^{2}.$ Only one variable is squared. $x=3{y}^{2}-2y+1$ **Circle**${x}^{2}\text{-}$ and ${y}^{2}\text{-}$ terms have the same coefficients ${x}^{2}+{y}^{2}=49$ **Ellipse**${x}^{2}\text{-}$ and ${y}^{2}\text{-}$ terms have the **same**sign, different coefficients$4{x}^{2}+25{y}^{2}=100$ **Hyperbola**${x}^{2}\text{-}$ and ${y}^{2}\text{-}$ terms have **different**signs, different coefficients$25{y}^{2}-4{x}^{2}=100$

**How to solve a system of nonlinear equations by graphing.**- Step 1. Identify the graph of each equation. Sketch the possible options for intersection.
- Step 2. Graph the first equation.
- Step 3. Graph the second equation on the same rectangular coordinate system.
- Step 4. Determine whether the graphs intersect.
- Step 5. Identify the points of intersection.
- Step 6. Check that each ordered pair is a solution to both original equations.

**How to solve a system of nonlinear equations by substitution.**- Step 1.
Identify the graph of each equation. Sketch the possible options for intersection.
- Step 2. Solve one of the equations for either variable.
- Step 3. Substitute the expression from Step 2 into the other equation.
- Step 4. Solve the resulting equation.
- Step 5. Substitute each solution in Step 4 into one of the original equations to find the other variable.
- Step 6. Write each solution as an ordered pair.
- Step 7.
Check that each ordered pair is a solution to
**both**original equations.

- Step 1.
Identify the graph of each equation. Sketch the possible options for intersection.
**How to solve a system of equations by elimination.**- Step 1. Identify the graph of each equation. Sketch the possible options for intersection.
- Step 2. Write both equations in standard form.
- Step 3.
Make the coefficients of one variable opposites.

Decide which variable you will eliminate.

Multiply one or both equations so that the coefficients of that variable are opposites. - Step 4. Add the equations resulting from Step 3 to eliminate one variable.
- Step 5. Solve for the remaining variable.
- Step 6. Substitute each solution from Step 5 into one of the original equations. Then solve for the other variable.
- Step 7. Write each solution as an ordered pair.
- Step 8.
Check that each ordered pair is a solution to
**both**original equations.