To solve for x and y, set both equations equal to each other and solve for x.

\(\displaystyle -x^2+4=x^2+4\)

\(\displaystyle 2x^2 =0\)

\(\displaystyle x=0\)

Substitute \(\displaystyle x=0\) into either parabola.

\(\displaystyle y=0^2+4 = 4\)

The coordinate of intersection is \(\displaystyle (0,4)\).

Set both parabolas equal to each other and solve for x.

\(\displaystyle 2x^2+3x+4=-4x^2+4\)

\(\displaystyle 6x^2+3x=0\)

\(\displaystyle 3x(2x+1)=0\)

\(\displaystyle x=0,-\frac{1}{2}\)

Substitute both values of \(\displaystyle x\) into either parabola and determine \(\displaystyle y\).

\(\displaystyle y=-4(0)^2+4= 4\)

\(\displaystyle y=-4\left(-\frac{1}{2}\right)^2+4= -4\left(\frac{1}{4}\right)+4= -1+4=3\)

The coordinates of intersection are:

\(\displaystyle (0,4)\) and \(\displaystyle \left(-\frac{1}{2},3\right)\)

To solve, set both equations equal to each other:

\(\displaystyle x^2 + 40x - 6 = -x^2 + 19x + 5\)

To solve as a quadratic, combine like terms by adding/subtracting all three terms from the right side to the left side:

\(\displaystyle x^2 + x^2 +40x - 19x - 6 - 5 = 0\)

This simplifies to

\(\displaystyle 2x^2 + 21 x - 11 = 0\)

Solving by factoring or the quadratic formula gives the solutions \(\displaystyle -11\) and \(\displaystyle 0.5\).

Plugging each into either original equation gives us:

\(\displaystyle (-11)^2 + 40(-11 ) - 6 = 121 - 440 - 6 = -325\)

\(\displaystyle (0.5) ^ 2 + 40 (0.5) - 6 = 14.25\)

Our coordinate pairs are \(\displaystyle (-11, -325)\) and \(\displaystyle (0.5, 14.25 )\).

To solve, set the two quadratics equal to each other and then combine like terms:

\(\displaystyle 5 x ^ 2 - 10 x + 6 = 2x ^2 + 9x + 20\) subtract everything on the right from both sides to combine like terms.

\(\displaystyle 5x^2 - 2x^2 - 10x - 9 x +6 - 20 = 0\)

\(\displaystyle 3x^2 - 19x - 14=0\)

Solving by factoring or using the quadratic formula gives us the solutions \(\displaystyle -\frac{2}{3}\) and \(\displaystyle 7\).

To find the y-coordinates, plug these into either equation:

\(\displaystyle 5\left( -\frac{ 2}{3} \right)^2 -10 \left(-\frac{ 2}{3}\right) + 6 = 14.\overline{8}\)

\(\displaystyle 2(7)^2 + 9(7) + 20 = 181\)

To solve, first re-write the second one so that y is isolated on the left side:

\(\displaystyle y = -2x ^ 2 - 18 x + 10\)

Now set the two quadratics equal to each other:

\(\displaystyle x^ 2 - 7 x - 10 = -2x^2 - 18x + 10\) add/subtract all of the terms from the right side so that this is a quadratic equal to zero.

\(\displaystyle x^2 + 2x^2 - 7x + 18 x -10 -10 = 0\) combine like terms.

\(\displaystyle 3x^2 + 11x -20 = 0\)

Using the quadratic formula or by factoring, we get the two solutions \(\displaystyle -5\) and \(\displaystyle \frac{ 4}{3}\).

To get the y-coordinates, plug these numbers into either function:

\(\displaystyle y = (-5)^2 - 7 (-5) - 10 = 25+35 - 10 = 50\)

\(\displaystyle y = \left(\frac{4}{3}\right)^2 - 7\left(\frac{ 4}{3} \right) - 10 = \frac {16}{9} - \frac{84}{9} - \frac{90}{9} = \frac{-158}{9 } = -17.\overline{5}\)

To solve, set the two polynomials equal to each other:

\(\displaystyle 4x^3 + 3x^2 -10x - 5 = 4x^3 + 2x^2 -8x + 10\) add/subtract all of the terms from the right side from both sides.

\(\displaystyle 4x^3 - 4x^3 + 3x^2 - 2x^2 -10x + 8 x - 5 -10 = 0\) combine like terms.

\(\displaystyle x^2 -2x -15 = 0\)

Solving with the quadratic formula or by factoring gives us the solutions 5 and -3.

To get the y-coordinates, plug these numbers into either of the original equations:

\(\displaystyle 4(5)^3 +3(5)^2 -10(5) - 5 = 500 + 75 - 50 - 5 = 520\)

\(\displaystyle 4(-3)^3 + 2(-3) ^ 2 - 8 (-3 ) + 10 = -108 + 18 + 24 + 10 = -56\)

Recall that the polar equations of conic sections can come in the following forms:

\(\displaystyle r=\frac{ed}{1\pm e \cos(\theta)}\)

\(\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}\), where \(\displaystyle e\) is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

\(\displaystyle 0< e< 1\) will give an ellipse.

\(\displaystyle e=1\) will give a parabola.

\(\displaystyle e>1\) will give a hyperbola.

Now, for the given conic section, \(\displaystyle e>1\) so it must be a hyperbola.

Recall that the polar equations of conic sections can come in the following forms:

\(\displaystyle r=\frac{ed}{1\pm e \cos(\theta)}\)

\(\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}\), where \(\displaystyle e\) is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

\(\displaystyle 0< e< 1\) will give an ellipse.

\(\displaystyle e=1\) will give a parabola.

\(\displaystyle e>1\) will give a hyperbola.

Now, for the given conic section, \(\displaystyle 0< e< 1\) so it must be an ellipse.

Recall that the polar equations of conic sections can come in the following forms:

\(\displaystyle r=\frac{ed}{1\pm e \cos(\theta)}\)

\(\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}\), where \(\displaystyle e\) is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

\(\displaystyle 0< e< 1\) will give an ellipse.

\(\displaystyle e=1\) will give a parabola.

\(\displaystyle e>1\) will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by \(\displaystyle 4\).

\(\displaystyle r=\frac{8}{4-\sin\theta}=\frac{2}{1-\frac{1}{4}\sin\theta}\)

Now, for the given conic section, \(\displaystyle 0< e< 1\) so it must be an ellipse.

Recall that the polar equations of conic sections can come in the following forms:

\(\displaystyle r=\frac{ed}{1\pm e \cos(\theta)}\)

\(\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}\), where \(\displaystyle e\) is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

\(\displaystyle 0< e< 1\) will give an ellipse.

\(\displaystyle e=1\) will give a parabola.

\(\displaystyle e>1\) will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by \(\displaystyle 6\).

\(\displaystyle r=\frac{6}{6-6\cos\theta}=\frac{1}{1-\cos\theta}\)

Now, for the given conic section, \(\displaystyle e=1\) so it must be a parabola.