Question 2.6.3: Express y=2x²-6x+4 as a standard equation of a parabola with......

Precalculus Functions and Graphs [1893897]

Express y=2x²-6x+4 as a standard equation of a parabola with a vertical axis. Find the vertex and sketch the graph.

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Equation: y = 2x² – 6x + 4

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Step 1:

Factor out the common factor of 2 from the first two terms: y = 2(x^2 - 3x) + 4.

Step 2:

To complete the square, we need to add and subtract the square of half the coefficient of the x-term. In this case, the coefficient of the x-term is -3, so we add and subtract (-3/2)^2 = 9/4 inside the parentheses: y = 2(x^2 - 3x + 9/4 - 9/4) + 4.

Step 3:

Simplify inside the parentheses: y = 2((x - 3/2)^2 - 9/4) + 4.

Step 4:

Distribute the 2 to both terms inside the parentheses: y = 2(x - 3/2)^2 - 2(9/4) + 4.

Step 5:

Simplify the expression: y = 2(x - 3/2)^2 - 9/2 + 4.

Step 6:

Combine like terms: y = 2(x - 3/2)^2 - 1/2.

Step 7:

The final equation is in vertex form, y = a(x - h)^2 + k, where the vertex is located at the point (h, k). In this case, the vertex is (3/2, -1/2).

Step 8:

Since the coefficient of the x^2 term, a, is positive (a = 2 > 0), the parabola opens upward.

Step 9:

To find the y-intercept, we substitute x = 0 into the equation: y = 2(0)^2 - 6(0) + 4 = 4. So the y-intercept is 4.

Step 10:

To find the x-intercepts, we set y = 0 and solve for x. We have the equation 2x^2 - 6x + 4 = 0. Factoring this equation or using the quadratic formula, we find that the x-intercepts are x = 1 and x = 2.

Final Answer

\begin{aligned}y & =2 x^2-6 x+4 & & \text { given } \\& =2\left(x^2-3 x+\right)+4 & & \text { factor out } 2 \text { from } 2 x^2-6 x \\& =2\left(x^2-3 x+\frac{9}{4}\right)+\left(4-\frac{9}{2}\right) & & \text { complete the square for } x^2-3 x \\& =2\left(x-\frac{3}{2}\right)^2-\frac{1}{2} & & \text { equivalent equation }\end{aligned}

The last equation has the form of the standard equation of a parabola with a=2, $h=\frac{3}{2}, \text {and} k=-\frac{1}{2}$ . Hence, the vertex V(h, k) of the parabola is $V\left(\frac{3}{2},-\frac{1}{2}\right)$ . Since a=2>0, the parabola opens upward.

To find the y-intercept of the graph of y=2x²-6x+4, we let x=0, obtaining y=4. To find the x-intercepts, we let y=0 and solve the equation 2x²-6x+4=0 or the equivalent equation 2(x-1)(x-2)=0, obtaining x=1 and x=2. Plotting the vertex and using the x – and y-intercepts provides enough points for a reasonably accurate sketch (see Figure 5).