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graph the equation. y = 2x² - 8x + 3

Question

graph the equation.
y = 2x² - 8x + 3

Explanation:

Step1: Find the vertex form

The quadratic equation is \( y = 2x^2 - 8x + 3 \). First, factor out the coefficient of \( x^2 \) from the first two terms: \( y = 2(x^2 - 4x) + 3 \). Then complete the square inside the parentheses. The formula for completing the square for \( x^2 + bx \) is \( (x + \frac{b}{2})^2 - (\frac{b}{2})^2 \). Here, \( b = -4 \), so \( \frac{b}{2} = -2 \), and \( (\frac{b}{2})^2 = 4 \). So we add and subtract 4 inside the parentheses: \( y = 2(x^2 - 4x + 4 - 4) + 3 \). This can be rewritten as \( y = 2((x - 2)^2 - 4) + 3 \). Distribute the 2: \( y = 2(x - 2)^2 - 8 + 3 \), which simplifies to \( y = 2(x - 2)^2 - 5 \).

Step2: Identify the vertex

From the vertex form \( y = a(x - h)^2 + k \), the vertex is \( (h, k) \). So for \( y = 2(x - 2)^2 - 5 \), the vertex is \( (2, -5) \).

Step3: Find the y - intercept

To find the y - intercept, set \( x = 0 \) in the original equation \( y = 2(0)^2 - 8(0) + 3 \), so \( y = 3 \). So the y - intercept is \( (0, 3) \).

Step4: Find the axis of symmetry

The axis of symmetry of a parabola in the form \( y = a(x - h)^2 + k \) is \( x = h \). So from the vertex form, the axis of symmetry is \( x = 2 \).

Step5: Plot the points and draw the parabola

  • Plot the vertex \( (2, -5) \).
  • Plot the y - intercept \( (0, 3) \). Since the parabola is symmetric about \( x = 2 \), the point symmetric to \( (0, 3) \) with respect to \( x = 2 \) is \( (4, 3) \) (because the distance from \( x = 0 \) to \( x = 2 \) is 2, so we move 2 units to the right of \( x = 2 \) to get \( x = 4 \)).
  • Then, using the shape of the parabola (since \( a = 2>0 \), the parabola opens upwards), draw a smooth curve through these points.

Answer:

To graph \( y = 2x^2 - 8x + 3 \):

  1. Vertex: \( (2, -5) \) (from vertex form \( y = 2(x - 2)^2 - 5 \)).
  2. Y - intercept: \( (0, 3) \) (set \( x = 0 \), \( y = 3 \)).
  3. Axis of Symmetry: \( x = 2 \).
  4. Symmetric Point to Y - intercept: \( (4, 3) \) (due to symmetry about \( x = 2 \)).
  5. Draw a parabola opening upwards (since \( a = 2>0 \)) through the vertex \( (2, -5) \), \( (0, 3) \), and \( (4, 3) \).