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Question: How do you write \(y=-3{{x}^{2}}+5x-2\) in vertex form?...

How do you write y=−3x2+5x−2y=-3{{x}^{2}}+5x-2 in vertex form?

Explanation

Solution

The vertex form of the equation of the form y=ax2+bx+cy=a{{x}^{2}}+bx+c is given by y=a(x−h)2+k.y=a{{\left( x-h \right)}^{2}}+k. In this equation, hh is the x−x-coordinate of the vertex. Also, kk is the y−y-coordinate of the vertex.

Complete step by step solution:
Consider the equation of the form y=ax2+bx+c.y=a{{x}^{2}}+bx+c.
The vertex form of the above equation is given by y=a(x−h)2+ky=a{{\left( x-h \right)}^{2}}+k where aa is the coefficient of x2,{{x}^{2}}, hh is the x−x-coordinate of the vertex and kk is the y−y-coordinate of the vertex.
In this equation, hh is obtained by h=−b2ah=\dfrac{-b}{2a} and kk is obtained by applying the value of hh in the above equation for kk is the y−y-coordinate corresponding to the x−x-coordinate h.h.
Let us consider the given equation y=−3x2+5x−2.y=-3{{x}^{2}}+5x-2.
We are asked to find the vertex form of the given equation.
So, let us write the coefficients first.
The coefficient of the term x2{{x}^{2}} is −3.-3. That is, a=−3.a=-3.
The coefficient of the term xx is 5.5. That is b=5.b=5.
Now, we are going to find the value of the x−x-coordinate.
When we apply the values in the formula for h,h, we will get h=−b2a=−52⋅(−3).h=\dfrac{-b}{2a}=\dfrac{-5}{2\cdot \left( -3 \right)}.
Now we will get h=−5−6=56.h=\dfrac{-5}{-6}=\dfrac{5}{6}. Therefore, the value of x−x-coordinate of the vertex is h=56.h=\dfrac{5}{6}.
We are going to apply this value in the given equation to get the y−y-coordinate kk of the vertex.
So, k=−3(56)2+5(56)−2.k=-3{{\left( \dfrac{5}{6} \right)}^{2}}+5\left( \dfrac{5}{6} \right)-2.
Now we will get k=−32536+556−2.k=-3\dfrac{25}{36}+5\dfrac{5}{6}-2.
That is, k=−32536+256−2.k=-3\dfrac{25}{36}+\dfrac{25}{6}-2.
Now, cancel the common factor 33 in the first summand from the numerator and the denominator to get k=−2512+256−2.k=\dfrac{-25}{12}+\dfrac{25}{6}-2.
Make the denominators the same, k=−2512+2×252×6−2×1212=−2512+5012−2412k=\dfrac{-25}{12}+\dfrac{2\times 25}{2\times 6}-\dfrac{2\times 12}{12}=\dfrac{-25}{12}+\dfrac{50}{12}-\dfrac{24}{12}
Now we will get k=2512−2412=112.k=\dfrac{25}{12}-\dfrac{24}{12}=\dfrac{1}{12}.
Now the vertex form of the equation is obtained by substituting these values in the equation y=a(x−h)2+ky=a{{\left( x-h \right)}^{2}}+k as, y=−3(x−56)2+112.y=-3{{\left( x-\dfrac{5}{6} \right)}^{2}}+\dfrac{1}{12}.
Hence the vertex form is y=−3(x−56)2+112.y=-3{{\left( x-\dfrac{5}{6} \right)}^{2}}+\dfrac{1}{12}.

Note: We can simplify the vertex form y=−3(x−56)2+112y=-3{{\left( x-\dfrac{5}{6} \right)}^{2}}+\dfrac{1}{12} as follows:
Now we get y=−3(x2−2x56+(56)2)+112,y=-3\left( {{x}^{2}}-2x\dfrac{5}{6}+{{\left( \dfrac{5}{6} \right)}^{2}} \right)+\dfrac{1}{12}, since (a−b)2=a2−2ab+b2.{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}.
Now we are cancelling the common factors from the numerator and the denominator of the second term and squaring the third term, y=−3(x2−53x+2536)+112.y=-3\left( {{x}^{2}}-\dfrac{5}{3}x+\dfrac{25}{36} \right)+\dfrac{1}{12}.
Take −3-3 inside the bracket as y=−3x2+353−32536+112.y=-3{{x}^{2}}+3\dfrac{5}{3}-3\dfrac{25}{36}+\dfrac{1}{12}.
We will get y=−3x2+5x−2512+112=−3x2+5x−2412.y=-3{{x}^{2}}+5x-\dfrac{25}{12}+\dfrac{1}{12}=-3{{x}^{2}}+5x-\dfrac{24}{12}.