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Question: How do you graph \[y = 3{\left( {x + 3} \right)^2} - 3\] ?...

How do you graph y=3(x+3)23y = 3{\left( {x + 3} \right)^2} - 3 ?

Explanation

Solution

Hint : Here in this question, we have to plot a graph for the given equation. The given equation resembles or recognised as the standard equation for a parabola, which is f(x)=a(xh)2+kf\left( x \right) = a{\left( {x - h} \right)^2} + k , where (h,k)\left( {h,k} \right) is the vertex of the parabola and next find the focus, directrix and the points to plot the required graph.

Complete step-by-step answer :
We know, in the quadratic equation f(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c , a b and c are the constants and xx is the variable. So, by finding the different values of xx and corresponding values of yy or f(x)f\left( x \right) , we can plot all the points in the graph and by joining all of them we can get the required shape.
here Parabola Formula for the equation of a parabola given in its standard form f(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c is:
Vertex of Parabola = (b2a,4acb24a)\left( {\dfrac{{ - b}}{{2a}},\dfrac{{4ac - {b^2}}}{{4a}}} \right)
Focus of the parabola = (b2a,4acb2+14a)\left( {\dfrac{{ - b}}{{2a}},\dfrac{{4ac - {b^2} + 1}}{{4a}}} \right)
Directrix of parabola y=4acb214ay = \dfrac{{4ac - {b^2} - 1}}{{4a}}
Now Consider, the given equation
y=3(x+3)23\Rightarrow \,y = 3{\left( {x + 3} \right)^2} - 3
On expanding (x+3)2{\left( {x + 3} \right)^2} using identities (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab
y=3(x2+32+2(x)(3))3\Rightarrow \,y = 3\left( {{x^2} + {3^2} + 2\left( x \right)\left( 3 \right)} \right) - 3
y=3(x2+9+6x)3\Rightarrow \,y = 3\left( {{x^2} + 9 + 6x} \right) - 3
y=3x2+27+18x3\Rightarrow \,y = 3{x^2} + 27 + 18x - 3
On simplification, we get
y=3x2+18x+24\Rightarrow \,y = 3{x^2} + 18x + 24
Where, a=3a = 3 , b=18b = 18 and c=24c = 24
Vertex of Parabola = (b2a,4acb24a)\left( {\dfrac{{ - b}}{{2a}},\dfrac{{4ac - {b^2}}}{{4a}}} \right)
(182(3),4(3)(24)(18)24(3))\Rightarrow \,\,\left( {\dfrac{{ - 18}}{{2\left( 3 \right)}},\dfrac{{4\left( 3 \right)\left( {24} \right) - {{\left( {18} \right)}^2}}}{{4\left( 3 \right)}}} \right)
(186,28832412)\Rightarrow \,\,\left( {\dfrac{{ - 18}}{6},\dfrac{{288 - 324}}{{12}}} \right)
(3,3)\Rightarrow \,\,\left( { - 3, - 3} \right)
Vertex = (3,3)\left( { - 3, - 3} \right)
Focus of the parabola = (b2a,4acb2+14a)\left( {\dfrac{{ - b}}{{2a}},\dfrac{{4ac - {b^2} + 1}}{{4a}}} \right)
(182(3),4(3)(24)(18)2+14(3))\Rightarrow \,\,\left( {\dfrac{{ - 18}}{{2\left( 3 \right)}},\dfrac{{4\left( 3 \right)\left( {24} \right) - {{\left( {18} \right)}^2} + 1}}{{4\left( 3 \right)}}} \right)
(186,288324+112)\Rightarrow \,\,\left( {\dfrac{{ - 18}}{6},\dfrac{{288 - 324 + 1}}{{12}}} \right)
(186,36+112)\Rightarrow \,\,\left( {\dfrac{{ - 18}}{6},\dfrac{{ - 36 + 1}}{{12}}} \right)
(186,3512)\Rightarrow \,\,\left( {\dfrac{{ - 18}}{6},\dfrac{{ - 35}}{{12}}} \right)
Focus = (186,3512)\left( {\dfrac{{ - 18}}{6},\dfrac{{ - 35}}{{12}}} \right)
Directrix of parabola y=4acb214ay = \dfrac{{4ac - {b^2} - 1}}{{4a}}
4(3)(24)(18)214(3)\Rightarrow \,\,\dfrac{{4\left( 3 \right)\left( {24} \right) - {{\left( {18} \right)}^2} - 1}}{{4\left( 3 \right)}}
288324112\Rightarrow \,\,\dfrac{{288 - 324 - 1}}{{12}}
36112\Rightarrow \,\,\dfrac{{ - 36 - 1}}{{12}}
3712\Rightarrow \,\,\dfrac{{ - 37}}{{12}}
Directrix y=3712y = \dfrac{{ - 37}}{{12}}
Now putting in different values for x in the equation y=3(x+3)23y = 3{\left( {x + 3} \right)^2} - 3 and calculate the corresponding values for y.
When x=5y=3(5+3)23y=9x = - 5 \Rightarrow y = 3{\left( { - 5 + 3} \right)^2} - 3 \Rightarrow y = 9
When x=4y=3(4+3)23y=0x = - 4 \Rightarrow y = 3{\left( { - 4 + 3} \right)^2} - 3 \Rightarrow y = 0
When x=3y=3(3+3)23y=3x = - 3 \Rightarrow y = 3{\left( { - 3 + 3} \right)^2} - 3 \Rightarrow y = - 3
When x=2y=3(2+3)23y=0x = - 2 \Rightarrow y = 3{\left( { - 2 + 3} \right)^2} - 3 \Rightarrow y = 0
When x=1y=3(1+3)23y=9x = - 1 \Rightarrow y = 3{\left( { - 1 + 3} \right)^2} - 3 \Rightarrow y = 9

xx5 - 54 - 43 - 32 - 21 - 1
yy99003 - 30099
(x,y)\left( {x,y} \right)(5,9)\left( { - 5,9} \right)(4,0)\left( { - 4,0} \right)(3,3)\left( { - 3, - 3} \right)(2,0)\left( { - 2,0} \right)(1,9)\left( { - 1,9} \right)

Now, plot the required graph:
we determine the values and finally we obtain the linear equation for y and hence we can plot the graph.

Note : The plotting of a graph is different here. When we plot the graph for the parabola we use different methods. Here in this question Vertex of Parabola = (b2a,4acb24a)\left( {\dfrac{{ - b}}{{2a}},\dfrac{{4ac - {b^2}}}{{4a}}} \right) , Focus of the parabola = (b2a,4acb2+14a)\left( {\dfrac{{ - b}}{{2a}},\dfrac{{4ac - {b^2} + 1}}{{4a}}} \right) , Directrix of parabola y=4acb214ay = \dfrac{{4ac - {b^2} - 1}}{{4a}}