Marvin has a coupon that discounts the rental of a full-size car by $25. the rental car costs $49 per

Marvin has a coupon that discounts the rental of a full-size car by $25. the rental car costs $49 per day and the insurance $21 per day. if the cost is $465, how many days will the rent the car for? write and solve the equation

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  1. 1. f(x)=x²+10x+16

    use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(as, a> 0 the parabola is open upward), b=10. by putting the values.

    -b/2a = -10/2(1) = -5

    f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

    so, vertex = (-5, -9)

    now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

    now find x-intercept put y=0 in the above equation. 0= x²+10x+16

    x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

    from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

    2. f(x)=−(x−3)(x+1)

    by multiplying the factors, the general form is f(x)= -x²+2x+3.

    use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=2. by putting the values.

    -b/2a = -2/2(-1) = 1

    f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

    so, vertex = (1, 4)

    now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

    now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

    -x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

    from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

    3. f(x)= −x²+4

    use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=0. by putting the values.

    -b/2a = -0/2(-1) = 0

    f(-b/2a)= f(0)= −(0)²+4 =4

    so, vertex = (0, 4)

    now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

    now find x-intercept put y=0 in the above equation. 0= −x²+4

    −x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

    from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

    4. f(x)=2x²+16x+30

    use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(as, a> 0 the parabola is open upward), b=16. by putting the values.

    -b/2a = -16/2(2) = -4

    f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

    so, vertex = (-4, -2)

    now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

    now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

    2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

    from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

    5. y=(x+2)²+4

    the general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

    if a> 0 parabola is opened upward.

    if a< 0 parabola is opened downward.

    compare the given equation with general form of parabola.

    -h=2 ⇒h=-2

    k=4

    so, vertex= (-2, 4)

    as, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.

    hope this : )

    [tex]Use the parabola tool to graph the quadratic function f(x) = -2(x+4)2 - 3 graph the parabola by firs[/tex]

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