Question #261683

Solve the equation 4x-3 = 2x2 + 5x -6. Please indicate what you graphed and what you got the calculator to do.


1
Expert's answer
2021-11-08T13:17:21-0500

The graph of the quadratic function y=2x2+5x6y=2x^2+5x-6 is parabola.


y=2x2+5x6y=2x^2+5x-6

=2(x2+2(54)x+(54)2)2(54)26=2\bigg(x^2+2(\dfrac{5}{4})x+(\dfrac{5}{4})^2\bigg)-2(\dfrac{5}{4})^2-6

=2(x+54)2738=2\big(x+\dfrac{5}{4}\big)^2-\dfrac{73}{8}

Vertex:(54,738)Vertex:\big(-\dfrac{5}{4}, -\dfrac{73}{8}\big)

yy -intersection: x=0,y=2(0)2+5(0)6=6x=0, y=2(0)^2+5(0)-6=-6

Point (0,6)(0, -6)


xx -intersection(s): y=0,0=2(x+54)2738y=0, 0=2\big(x+\dfrac{5}{4}\big)^2-\dfrac{73}{8}

2(x+54)2=7382\big(x+\dfrac{5}{4}\big)^2=\dfrac{73}{8}

(x+54)2=7316\big(x+\dfrac{5}{4}\big)^2=\dfrac{73}{16}

x+54=±7316x+\dfrac{5}{4}=\pm\sqrt{\dfrac{73}{16}}

x=54±734x=-\dfrac{5}{4}\pm\dfrac{\sqrt{73}}{4}

x1=54734x_1=-\dfrac{5}{4}-\dfrac{\sqrt{73}}{4}

x2=54+734x_2=-\dfrac{5}{4}+\dfrac{\sqrt{73}}{4}

Point (5734,0),\big(\dfrac{-5-\sqrt{73}}{4}, 0\big), Point (5+734,0).\big(\dfrac{-5+\sqrt{73}}{4}, 0\big).


x=1:y=2(1)2+5(1)6=1x=1: y=2(1)^2+5(1)-6=1

Point (1,1).(1, 1).


x=2:y=2(2)2+5(2)6=8x=-2: y=2(-2)^2+5(-2)-6=-8

Point (1,9).(-1, -9).


The graph of the linear function y=4x3y=4x-3 is a straight line.

x=0:y=4(0)3=3x=0: y=4(0)-3=-3 .

Point (0,3)(0, -3)


y=0:0=4x3=>x=34y=0: 0=4x-3=>x=\dfrac{3}{4}

Point (34,0)\big(\dfrac{3}{4}, 0\big)

Points of intersection (1,1),(1.5,9)(1, 1), (-1.5, -9)

Check


2(1)2+5(1)6=4(1)32(1)^2+5(1)-6=4(1)-3

1=1,True1=1, True


2(1.5)2+5(1.5)6=4(1.5)32(-1.5)^2+5(-1.5)-6=4(-1.5)-3

9=9,True-9=-9, True



Solve the equation algebraically


2x2+5x6=4x32x^2+5x-6=4x-3

2x2+5x64x+3=02x^2+5x-6-4x+3=0

2x2+x3=02x^2+x-3=0


D=(1)24(2)(3)=25D=(1)^2-4(2)(-3)=25

x=1±252(2)=1±54x=\dfrac{-1\pm\sqrt{25}}{2(2)}=\dfrac{-1\pm5}{4}

x1=154=1.5x_1=\dfrac{-1-5}{4}=-1.5

y1=4(1.5)3=9y_1=4(-1.5)-3=-9

x2=1+54=1x_2=\dfrac{-1+5}{4}=1

y2=4(1)3=1y_2=4(1)-3=1

x1=1.5,x2=1x_1=-1.5, x_2=1


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