Graphing

Self-Test:

 1)       The circle given by $(x-7)^2+ (y+8)^2 = 25$ has: centre $(7,-8)$ and radius $25$ centre $(7,-8)$ and radius $5$ centre $(7,8)$ and radius $5$ centre $(-7,8)$ and radius $12.5$ centre $(7,-8)$ and radius $12.5$

Hint The equation of a circle is $(x-x_C)^2+(y-y_C)^2=r^2$, where $r$ is the radius and $(x_C,y_C)$ is the centre.

 2)       Let $f(x)=3x^2-2x+1$ and $g(x)=3x^2-2x-1$. Then: the graph of $y=f(x)$ has no roots; the graph of $y=g(x)$ has two distinct roots; and the two parabola vertices have the same $x$-coordinate the graph of $y=f(x)$ has two distinct roots; the graph of $y=g(x)$ has no roots; and the two parabola vertices have the same $x$-coordinate the graphs of both functions each have two distinct roots; but the parabola vertices have different $x$-coordinates the graph of $y=f(x)$ has no roots; the graph of $y=g(x)$ has two distinct roots; but the parabola vertices have different $x$-coordinates both graphs have exactly one distinct root, but the parabola vertices have different $x$-coordinates.

Hint 1 For a quadratic, $y=ax^2+bx+c$, the sign of the discriminant $D=b^2-4ac$ tells us the number of roots.
Hint 2 If we rewrite the equation in the form $y=a(x-h)^2+k$, the vertex is $(h,k)$.

 3)       To transform the graph of $y = x^2$ into the graph of $y = 5(x - 3)^2 + 2$, one must: stretch the graph in the vertical direction by a factor of $5$, shift to the left $3$ units and up two units. stretch the graph in the horizontal direction by a factor of $5$, shift to the left $3$ units and up two units. stretch the graph in the vertical direction by a factor of $5$, shift to the right $3$ units and up $2$ units. stretch the graph in the vertical direction by a factor of $5$, shift to the right $3$ units and down $2$ units. stretch the graph in the horizontal direction by a factor of $5$, shift to the right $3$ units and down $2$ units.

Hint Transforming $y=f(x)$ to $y=af(x+b)+c$, $a$ represents a vertical stretch/shrink, and $b$ and $c$ - a horizontal and a vertical shift respectively.

 4)       The graph of $y = f(x) = 1/x$ is the same as the graph of: $y = -f(x)$ $y = f(-x)$ $y = -f(-x)$ all of the above functions not any of the above functions

Hint Transforming $y=f(x)$ to $y=f(-x)$ and to $y=-f(x)$ means reflectinng in the $y$-axis and the $x$-axis respectively.

 5)       The hyperbola $\frac{x^2}{5}-\frac{y^2}{3}=1$ has foci: $(0,\pm 5)$ $(\pm 4,0)$ $(\pm 2\sqrt{2},0)$ $(0, \pm 2\sqrt{2})$ $(0,\pm 3)$

Hint The loci of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are given by $(\pm c, 0)$ where $c^2=a^2+b^2$.

 6)       The vertex of the parabola $y = x^2 + 2x - 15$ lies on the line: $y = 3$ $x = 3$ $x = -1$ $y = 1$ $y = -1$

Hint Rewriting the equation in the form: $y=a(x-h)^2+k$, the vertex is given by $(h,k)$.

 7)       Consider the conic sections $x^2+y^2=4$ and $\frac{x^2}{9}+\frac{y^2}{4}=1$. It is true that the circle is inscribed inside the ellipse. the ellipse is inscribed inside the circle. the graph of the circle is strictly inside the graph of the ellipse in some places and strictly outside in others. the ellipse is longer than the circle in the vertical direction. the circle is larger than the ellipse in the vertical direction.

Hint 1 In the equation of a circle $x^2+y^2=r^2$, the radius is given by $r$.
Hint 2 For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the axes are given by $(\pm a, 0)$ and $(0,\pm b)$.

 8)       The equation of the ellipse with foci $(0,\pm 2)$ and major axis of length $\sqrt{6}$ is: $\frac{x^2}{3} + \frac{y^2}{2}=1$ $\frac{x^2}{2} + \frac{y^2}{6}=1$ $\frac{x^2}{6} + \frac{y^2}{2}=1$ $\frac{x^2}{3} + \frac{y^2}{6}=1$ $\frac{x^2}{2} + \frac{y^2}{3}=1$

Hint For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the axes are given by $(\pm a, 0)$ and $(0,\pm b)$ and the foci are determined by $c$, where $c^2=|a^2-b^2|$.

 9)       The circle given by $3x^2-12x-18y+3y^2=0$ has radius equal to $13$ $3$ $\sqrt{39}$ $\sqrt{3}$ $\sqrt{13}$

Hint For a circle with equation $(x-x_C)^2+(y-y_C)^2=r^2$, the radius is $r$.

 10)       A parabola with equation $y = a(x - r)(x - s)$ has its vertex on the $y$-axis if and only if: $r × s = 0$. $r + s = 0$. $r = s$. $a$ is positive. $a$ is negative.

Hint Rewriting the equation in the form: $y=a(x-h)^2+k$, the vertex is given by $(h,k)$, and lying on the $y$-axis means $h=0$.