Write the slope-intercept form.

\(\displaystyle y=mx+b\)

The point given the x-intercept of 6 is \(\displaystyle (6,0)\).

Substitute the point and the slope into the equation and solve for the y-intercept.

\(\displaystyle 0=(1)(6)+b\)

\(\displaystyle b=-6\)

Substitute the y-intercept back to the slope-intercept form to get your equation.

\(\displaystyle y=x-6\)

The equation of a vertical parabola, in vertex form, is

\(\displaystyle y = a(x-h)^{2}+k\),

where \(\displaystyle (h,k)\) is the vertex. Set \(\displaystyle h = -4,k=10\):

\(\displaystyle y = a[x-(-4)]^{2}+10\)

\(\displaystyle y = a(x+4)^{2}+10\)

To find \(\displaystyle a\), use the \(\displaystyle y\)-intercept, setting \(\displaystyle x=0,y= -2\):

 \(\displaystyle -2 = a(0+4)^{2}+10\)

\(\displaystyle -2 = a\cdot 4^{2}+10\)

\(\displaystyle -2 = 16a +10\)

\(\displaystyle -12 = 16a\)

\(\displaystyle a =- \frac{3}{4}\)

The equation, in vertex form, is \(\displaystyle y =- \frac{3}{4}(x+4)^{2}+10\); in standard form:

\(\displaystyle y = -\frac{3}{4}(x+4)^{2}+10\)

\(\displaystyle y =- \frac{3}{4}(x^{2} + 8x+16)+10\)

\(\displaystyle y = -\frac{3}{4} x^{2}- 6x-12+10\)

\(\displaystyle y = -\frac{3}{4} x^{2}- 6x-2\)

The equation of a vertical parabola, in vertex form, is

\(\displaystyle y = a(x-h)^{2}+k\),

where \(\displaystyle (h,k)\) is the vertex. Set \(\displaystyle h = -3,k=-6\):

\(\displaystyle y = a[x-(-3)]^{2}+ (-6)\)

\(\displaystyle y = a(x+3)^{2}-6\)

To find \(\displaystyle a\), use the known \(\displaystyle x\)-intercept, setting \(\displaystyle x= -7, y=0\):

\(\displaystyle 0 = a(-7+3)^{2}-6\)

\(\displaystyle 0 = a(-4)^{2}-6\)

\(\displaystyle 0 =16 a-6\)

\(\displaystyle 16a= 6\)

\(\displaystyle a = \frac{6}{16} = \frac{3}{8}\)

The equation, in vertex form, is \(\displaystyle y = \frac{3}{8}(x+3)^{2}-6\); in standard form:

\(\displaystyle y = \frac{3}{8}(x+3)^{2}-6\)

\(\displaystyle y = \frac{3}{8}(x^{2}+6x+9)-6\)

\(\displaystyle y = \frac{3}{8}x^{2}+ \frac{9}{4}x+ \frac{27}{8}-6\)

\(\displaystyle y = \frac{3}{8}x^{2}+ \frac{9}{4}x+ \frac{27}{8}- \frac{48}{8}\)

\(\displaystyle y = \frac{3}{8}x^{2}+ \frac{9}{4}x - \frac{21}{8}\)

If a vertical parabola has only one \(\displaystyle x\)-intercept, which here is \(\displaystyle (6,0)\), that point doubles as its vertex as well. 

The equation of a vertical parabola, in vertex form, is

\(\displaystyle y = a(x-h)^{2}+k\),

where \(\displaystyle (h,k)\) is the vertex. Set \(\displaystyle h = 6,k=0\):

\(\displaystyle y = a(x-6)^{2}+0\)

\(\displaystyle y = a(x-6)^{2}\)

To find \(\displaystyle a\), use the \(\displaystyle y\)-intercept, setting \(\displaystyle x= 0, y=4\):

\(\displaystyle 4= a(0-6)^{2}\)

\(\displaystyle 4= a( -6)^{2}\)

\(\displaystyle 4 = 36a\)

\(\displaystyle a = \frac{4}{36} = \frac{1}{9}\)

The equation, in vertex form, is \(\displaystyle y = \frac{1}{9}(x-6)^{2}\). In standard form:

\(\displaystyle y = \frac{1}{9}(x-6)^{2}\)

\(\displaystyle y = \frac{1}{9}(x^{2}-12+36)\)

\(\displaystyle y = \frac{1}{9} x^{2}-\frac{4}{3}x+4\)

The equation of a vertical parabola, in standard form, is

\(\displaystyle y = ax^{2}+ bx + c\)

for some real \(\displaystyle a,b,c\). 

\(\displaystyle c\) is the \(\displaystyle y\)-coordinate of the \(\displaystyle y\)-intercept, so \(\displaystyle c=6\), and the equation is

\(\displaystyle y = ax^{2}+ bx + 6\)

Set \(\displaystyle x=2, y=0\):

\(\displaystyle 0 = a\cdot 2^{2}+ b \cdot 2 + 6\)

\(\displaystyle 0 = 4a+2b + 6\)

However, no other information is given, so the values of \(\displaystyle a\) and \(\displaystyle b\) cannot be determined for certain. The correct response is that insufficient information is given.

The equation of the ellipse with center \(\displaystyle (h,k)\), horizontal axis of length \(\displaystyle 2a\), and vertical axis of length \(\displaystyle 2b\) is

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

The ellipse has center \(\displaystyle (6,2)\), horizontal axis of length 8, and vertical axis of length 16. Therefore,

\(\displaystyle h = 6,k = 2\), \(\displaystyle a = \frac{8}{2} = 4\), and \(\displaystyle b = \frac{16}{2} = 8\).

The equation of the ellipse is

\(\displaystyle \frac{(x-6)^{2}}{4^{2}} + \frac{(y-2)^{2}}{8^{2}} = 1\)

\(\displaystyle \frac{(x-6)^{2}}{16} + \frac{(y-2)^{2}}{64} = 1\)

The equation of the ellipse with center \(\displaystyle (h,k)\), horizontal axis of length \(\displaystyle 2a\), and vertical axis of length \(\displaystyle 2b\) is

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

The ellipse has center \(\displaystyle (1,1)\), horizontal axis of length 10, and vertical axis of length 6. Therefore,

\(\displaystyle h = k = 1\), \(\displaystyle a = \frac{10}{2} = 5\), and \(\displaystyle b = \frac{6}{2} = 3\).

The equation of the ellipse is

\(\displaystyle \frac{(x-1)^{2}}{5^{2}} + \frac{(y-1)^{2}}{3^{2}} = 1\)

\(\displaystyle \frac{(x-1)^{2}}{25} + \frac{(y-1)^{2}}{9} = 1\)

The equation of the ellipse with center \(\displaystyle (h,k)\), horizontal axis of length \(\displaystyle 2a\), and vertical axis of length \(\displaystyle 2b\) is

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

The ellipse has center \(\displaystyle (3,5)\), horizontal axis of length 8, and vertical axis of length 6. Therefore,

\(\displaystyle h = 3,k = 5\), \(\displaystyle a = \frac{8}{2} = 4\), and \(\displaystyle b = \frac{6}{2} = 3\).

The equation of the ellipse is

\(\displaystyle \frac{(x-3)^{2}}{4^{2}} + \frac{(y-5)^{2}}{3^{2}} = 1\)

\(\displaystyle \frac{(x-3)^{2}}{16} + \frac{(y-5)^{2}}{9} = 1\)

To find the \(\displaystyle y\)-intercept, that is, the point of intersection with the \(\displaystyle y\)-axis, of the line of equation \(\displaystyle 3x+ 5y = 30\), set \(\displaystyle x=0\) and solve for \(\displaystyle y\):

\(\displaystyle 3 (0)+ 5y = 30\)

\(\displaystyle 5y = 30\)

\(\displaystyle y = 6\)

The \(\displaystyle y\)-intercept is \(\displaystyle (0,6)\).

The \(\displaystyle x\)-intercept can be found by doing the opposite:

\(\displaystyle 3x+ 5 (0) = 30\)

\(\displaystyle 3x = 30\)

\(\displaystyle x=10\)

The \(\displaystyle x\)-intercept is \(\displaystyle (10,0)\).

The parabola has these intercepts as well. Also, since the vertical parabola has only one \(\displaystyle x\)-intercept, that point doubles as its vertex as well. 

The equation of a vertical parabola, in vertex form, is

\(\displaystyle y = a(x-h)^{2}+k\),

where \(\displaystyle (h,k)\) is the vertex. Set \(\displaystyle h = 10,k=0\):

\(\displaystyle y = a(x-10)^{2}+0\)

\(\displaystyle y = a(x-10)^{2}\)

for some real \(\displaystyle a\). To find it, use the \(\displaystyle y\)-intercept, setting \(\displaystyle x=0, y = 6\)

\(\displaystyle 6 = a(0-10)^{2}\)

\(\displaystyle 6 = a( -10)^{2}\)

\(\displaystyle 6 = 100a\)

\(\displaystyle a = \frac{6}{100} = \frac{3}{50}\)

The parabola has equation \(\displaystyle y =\frac{3}{50} (x-10)^{2}\), which is rewritten as

\(\displaystyle y =\frac{3}{50} (x^{2}-20x+100)\)

\(\displaystyle y = \frac{3}{50} x^{2}-\frac{6}{5}x+6\)

The equation of the ellipse with center \(\displaystyle (h,k)\), horizontal axis of length \(\displaystyle 2a\), and vertical axis of length \(\displaystyle 2b\) is

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

The center is \(\displaystyle (-4, 9)\), so \(\displaystyle h = -4\) and \(\displaystyle k = 9\).

To find \(\displaystyle a\), note that one endpoint of the horizontal axis is given by the point with the same \(\displaystyle y\)-coordinate through which it passes, namely, \(\displaystyle (11, 9)\). Half the length of this axis, which is \(\displaystyle a\), is the difference of the \(\displaystyle x\)-coordinates, so \(\displaystyle a = 11- (-4) = 15\). Similarly, to find \(\displaystyle b\), note that one endpoint of the vertical axis is given by the point with the same \(\displaystyle x\)-coordinate through which it passes, namely, \(\displaystyle (-4, 5)\). Half the length of this axis, which is \(\displaystyle b\), is the difference of the \(\displaystyle x\)-coordinates, so \(\displaystyle b = 9 - 5 = 4\).

The equation is 

\(\displaystyle \frac{(x-(-4))^{2}}{15^{2}} + \frac{(y-9)^{2}}{4^{2}} = 1\)

or 

\(\displaystyle \frac{(x+4)^{2}}{225} + \frac{(y-9)^{2}}{16} = 1\).