b) $(0,3)$ , $(-0.25,2.875)$

c) no solution

a)

$f(x)=-x^2+6x-5$ ; $g(x)=-4x+19$

$f(x)=g(x)$

$-x^2+6x-5=-4x+19$

$-x^2+10x-24=0$

$x^2-10x+24=0$

Solve by factoring

$$
(x-6)(x-4)=0
$$

$x=6$ and $x=4$

Substitute this values to either of the two equations. We shall choose $g(x)$

$g(6)=-4(6)+19=-5$

$g(4)=-4(4)+19=3$

Therefore, the points of intersection are $(6,-5)$ and $(4,3)$

$f(x)=2x^2-1$ ; $g(x)=3x+1$

$f(x)=g(x)$

$2x^2-1=3x+1$

$2x^2-3x-2=0$

Solve by factoring

$(x-2)(2x+1)=0$

$x=2$ and $x=-dfrac{1}{2}$

$g(2)=3(2)+1=7$

$g(-0.5)=3(-0.5)+1=-dfrac{1}{2}$

The points of intersection are $left(3,7right)$ and $left(-dfrac{1}{2},-dfrac{1}{2}right)$

$f(x)=3x^2-2x-1$ ; $g(x)=-x-6$

$f(x)=g(x)$

$3x^2-2x-1=-x-6$

$3x^2-x+5=0$

Remember that if $b^2-4ac<0$, there are no solutions

$a=3$ , $b=-1$, $c=5$

$b^2-4ac=(-1)^2-4(3)(5)=-59<0$

Thus, no intersection points.

b) (2,7), $left(-dfrac{1}{2},-dfrac{1}{2}right)$

c) no solution

Equate $f(x)$ and $g(x)$

$f(x)=g(x)$

$4x^2+x-3=5x-4$

$4x^2-4x+1=0$

If $b^2-4ac>0implies 2$ intersection points

$b^2-4ac=0implies$ only 1 intersection point

$b^2-4ac<0implies$ no intersection points

$a=4$, $b=-4$, $c=1$

$b^2-4ac=(-4)^2-4(4)(1)=0$

Since $b^2-4ac=0$, there is only one intersection point.

b) $(1.91,8.91)$ , $(-1.57,5.43)$

c) no solutions

d) $(-0.16,3.2)$, $(1.59,-3.95)$

“an integer two more than another integer” $implies$ $x+2$

“twice the larger integer is one more than the square of the smaller integer” $implies 2(x+2)=x^2+1$

$2x+4=x^2+1$

$x^2-2x-3=0$

Solve by factoring

$(x-3)(x+1)=0$

$x=3$ or $x=-1$

If $x=3$ , $x+2=5$

If $x=-1$ , $x+2=1$

Therefore, the integers that satisfy the conditions mentioned are

3 and 5 or $-1$ and $1$

Equate the revenue function and cost function

$R(t)=C(t)$

$-50t^2+300t=600-50t$

$50t^2-350t+600=0$

$50(t-4)(t-3)=0$

$t=4$ or $t=3$

Both values of $t$ makes $R(t)>0$ and $C(t)>0$, so the price that will allow it to break-even is

$$3.00$ or $$4.00$

(i) one point (red): $y=-4x+1$

(ii) two points (purple): $y=-4x+6$

(iii) no points (orange): $y=-4x-6$

The intersection point of the graph $f(x)=(x-2)^2-3$ and a line $g(x)=-4x+v$ occurs at values of $x$ where

$f(x)=g(x)$

$(x-2)^2-3=-4x+v$

Rewrite the equation in standard form.

$x^2-4x+4-3=-4x+v$

$x^2+1-v=0$

$a=1$, $b=0$ , $c=1-v$
We shall calculate the $b^2-4ac$ to know which values of $c$ can produce different number of solutions.

$b^2-4ac=0^2-4(1)(1-v)=-4+4v=4(v-1)$

2 intersection points: $4(v-1)>0implies v>1$

1 intersection point: $4(v-1)=0implies v=1$

no intersection point: $4(v-1)<0implies v<1$

Therefore, all lines with slope $-4$ that do not intersect with the given parabola should have $y$-intercepts less than 1.

b) (i) $y=-4x+1$, (ii) $y=-4x+6$ , (iii) $y=-4x-6$ ;

c) $y$-intercepts are all less than 1

$k=2x^2-8x+3$

$$
2x^2-8x+3-k=0
$$

$64-8(3-k)=0$

$64-24+8k=0$

$40+8k=0$

$8k=-40$

$$
k=-5
$$

$k=-3x^2-5x+4$

$$
-3x^2-5x+4-k=0
$$

$25+12(4-k)<0$

$25+48-12k<0$

$73-12k<0$

$-12kfrac{73}{12}
$$

$-4.9t^2+t+360=-4t+142$

$4.9t^2-5t-218=0$

We cannot factor this so we shall use quadratic formula

$t=dfrac{-bpmsqrt{b^2-4ac}}{2a}$

$t=dfrac{(-5)pmsqrt{(-5)^2-4(4.9)(-218)}}{2(4.9)}$

$t=-6.18$ or $7.20$

Choose the positive values.

$f(x)=g(x)$

$3x^2+4x-2=mx-5$

$3x^2+4x-2-mx+5=0$

$3x^2+x(4-m)+3=0$

$a=3$ , $b=(4-m)$ , $c=3$

$b^2-4ac=(4-m)^2-4(3)(3)=0$

$(4-m)^2=36$

$4-m=pm sqrt{36}$

$4-m=pm 6$

$m=4pm 6$

$m=-2$ or $10$

Therefore, the line with a $y$-intercept of $-5$ would be tangent to the parabola if it has slope of $-2$ or $10$.

$f(x)=g(x)$

$-4.9t^2+18.24t+0.8=-1.43t+4.26$

$4.9t^2-19.67t+3.46=0$

We shall use quadratic formula

$a=4.9$ , $b=-19.67$ , $c=3.46$

$t=dfrac{-bpmsqrt{b^2-4ac}}{2a}$

$t=dfrac{19.67pmsqrt{(-19.67)^2-4(4.9)(3.46)}}{2(4.9)}$

$t=3.83$ or $0.18$

If $t=3.83implies$ $g(3.83)=-1.43(3.83)+4.26=-1.21$ not possible since height must be positive

If $t=0.18implies$ $g(0.18)=-1.43(0.18)+4.26=4.00$ this is possible

Therefore, the blocker can block it after $0.18$ s at height 4 m above the ground

Note that the blocker’s hand must reach 4 m so he must jump very high. Theoretically it is possible, but practically it might be difficult.

1) Plot the graphs of each function then count the intersection points.

2) Equate the two functions and rewrite to standard form $ax^2+bx+c=0$. Then calculate $b^2-4ac$.

$b^2-4ac>0implies 2$ intersection points

$b^2-4ac=0implies$ 1 intersection point

$b^2-4ac<0implies$ no intersection points

(2) by calculating $b^2-4ac$

$2x+3y+6=0implies 3y=-2x-6$

Substitute this value of $3y$ to

$x^2-2x+3y+6=0$

$x^2-2x+(-2x-6)+6=0$

$x^2-4x=0$

$x(x-4)=0$

$x=0$ or $x=4$

Substitute this values of $x$ to the expression for $y$,
knowing that $3y=-2x-6$

$y=dfrac{3y}{3}=dfrac{-2x-6}{3}$

if $x=0implies y=dfrac{-2(0)-6}{3}=-2$

if $x=4implies y=dfrac{-2(4)-6}{3}=-dfrac{14}{3}$

Therefore, the intersection points are

$(0,-2)$ and $left(4,-dfrac{14}{3}right)$

$f(x)=g(x)$

$x^2-4=-3x^2+2x+8$

$4x^2-2x-12=0$

$2x^2-x-6=0$

$(x-2)(2x+3)=0$

$x=2$ or $x=-dfrac{3}{2}=-1.5$

We shall find the value of $y$ at each values of $x$ using either $f(x)$ or $g(x)$

$y=f(2)=2^2-4=0$

$y=f(-1.5)^2-4=-1.75$

Therefore, the points of intersection are

$(2,0)$ and $(-1.5,-1.75)$

Now we shall find the equation of the line that passes through these points

$m=dfrac{y_2-y_1}{x_2-x_1}=dfrac{-1.75-0}{-1.5-2}=0.5$

Now, we can write the point-slope form using either of the points

$y-y_1=m(x-x_1)$

$y-0=0.5(x-2)$

$y=0.5x-1$

Therefore, the equation of the line passing through the intersection points of $f(x)$ and $g(x)$ is

$$
y=0.5x-1
$$