Rewrite the equation in slope-intercept form, \(\displaystyle y=mx+b\).

\(\displaystyle -y-8=-6x+2\)

\(\displaystyle -y=-6x+10\)

\(\displaystyle y=6x-10\)

The y-intercept is \(\displaystyle b\), which is \(\displaystyle -10\).

The \(\displaystyle y\)-intercept of the graph of a function is the point at which it intersects the \(\displaystyle y\)-axis - that is, at which \(\displaystyle x = 0\). This point is \(\displaystyle \left (0, f(0) \right )\), so evaluate \(\displaystyle f(0)\):

\(\displaystyle f(x) = x^{2} + 3x - 10\)

\(\displaystyle f( 0) = 0^{2} + 3 \cdot 0 - 10 = 0 + 0 - 10 = -10\)

The \(\displaystyle y\)-intercept is \(\displaystyle \left (0, -10 \right )\).

The \(\displaystyle y\)-intercept of the graph of a function \(\displaystyle f\) has 0 as its \(\displaystyle x\)-coordinate, since it is defined to be the point at which it crosses the \(\displaystyle y\)-axis. Its \(\displaystyle y\)-coordinate is \(\displaystyle f(0)\), which can be found using substitution, as follows:

\(\displaystyle f(x) = x ^{2}+ 3x + 2\)

\(\displaystyle f(0) = 0 ^{2}+ 3 \cdot 0 + 2 = 0 + 0 + 2 = 2\)

The correct choice is \(\displaystyle (0,2)\).

An \(\displaystyle x\)-intercept of the graph of a function \(\displaystyle f\) has 0 as its \(\displaystyle y\)-coordinate, since it is defined to be a point at which it crosses the \(\displaystyle x\)-axis. Its \(\displaystyle x\)-coordinate is a value of \(\displaystyle x\) for which \(\displaystyle f(x) = 0\).

We can most easily determine whether \(\displaystyle (2, 0)\) is a point on the graph of \(\displaystyle f(x)\) by proving or disproving that \(\displaystyle f(2) = 0\), which we can do by substituting 2 for \(\displaystyle x\):

\(\displaystyle f(x) = x ^{2}+ 3x + 2\)

\(\displaystyle f(2) = 2^{2}+ 3 \cdot 2 + 2\)

\(\displaystyle f(2) = 4+ 6 + 2\)

\(\displaystyle f(2) = 1 2\)

\(\displaystyle f(2) \ne 0\), so \(\displaystyle (2, 0)\) is not an \(\displaystyle x\)-intercept. 

Similarly, substituting 3 for \(\displaystyle x\):

\(\displaystyle f(x) = x ^{2}+ 3x + 2\)

\(\displaystyle f(3) = 3^{2}+ 3 \cdot 3 + 2\)

\(\displaystyle f(3) = 9+ 9 + 2\)

\(\displaystyle f(3) = 1 2\)

\(\displaystyle f(3) \ne 0\), so \(\displaystyle (3, 0)\) is not an \(\displaystyle x\)-intercept. 

An \(\displaystyle x\)-intercept of the graph of a function \(\displaystyle f\) has 0 as its \(\displaystyle y\)-coordinate, since it is defined to be a point at which it crosses the \(\displaystyle x\)-axis. Its \(\displaystyle x\)-coordinate is a value of \(\displaystyle x\) for which \(\displaystyle f(x) = 0\), which can be found as follows:

Substituting the definition, we get 

\(\displaystyle 2x+ 7 = 0\)

Solving for \(\displaystyle x\) by subtracting 7 from both sides, then dividing both sides by 2:

\(\displaystyle 2x+ 7 -7 = 0 -7\)

\(\displaystyle 2x = -7\)

\(\displaystyle 2x \div 2 = -7 \div 2\)

\(\displaystyle x= - \frac{7}{2}\)

The \(\displaystyle x\)-intercept of the graph of \(\displaystyle f(x)\) is the point \(\displaystyle \left ( - \frac{7}{2}, 0 \right )\).

To determine which of the four choices is correct, substitute \(\displaystyle - \frac{7}{2}\) for \(\displaystyle x\) and determine for which definition of \(\displaystyle g\) it holds that \(\displaystyle g \left ( - \frac{7}{2} \right ) = 0\).

\(\displaystyle g(x) = 7\) can be eliminated immediately as a choice since it cannot take the value 0.

\(\displaystyle g(x) = 2x+ 14\)

\(\displaystyle g\left ( - \frac{7}{2} \right ) = 2\left ( - \frac{7}{2} \right ) + 14 = -7 + 14 = 7\)

\(\displaystyle g(x) = 4x+ 7\):

\(\displaystyle g \left ( - \frac{7}{2} \right ) = 4 \left ( - \frac{7}{2} \right ) + 7 = -14 + 7 = -7\)

\(\displaystyle g(x) = 4x+ 14\)

\(\displaystyle g \left ( - \frac{7}{2} \right ) = 4 \left ( - \frac{7}{2} \right ) + 14 = -14 + 14 =0\)

The correct choice is \(\displaystyle g(x) = 4x+ 14\).

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.