Questions: Assignment 5.1: Quadratic Functions Question 11 Fill in the info below. Then graph the parabola on the axes below by first clicking on the vertex, and then on another point close to the vertex that fits on the axes. y=x^2-10x+21 1. State whether the parabola opens up or down? Up Down 2. State the vertex as an ordered pair: 3. State the y-intercept as an ordered pair 4. Write the Equation of the Axis of Symmetry 5. Click on the vertex and then another point of the parabola that fits on the axes below. Clear All Draw:

Assignment 5.1: Quadratic Functions Question 11 Fill in the info below. Then graph the parabola on the axes below by first clicking on the vertex, and then on another point close to the vertex that fits on the axes. y=x^2-10x+21 1. State whether the parabola opens up or down? Up Down 2. State the vertex as an ordered pair: 3. State the y-intercept as an ordered pair 4. Write the Equation of the Axis of Symmetry 5. Click on the vertex and then another point of the parabola that fits on the axes below. Clear All Draw:
Transcript text: Assignment 5.1: Quadratic Functions Question 11 Fill in the info below. Then graph the parabola on the axes below by first clicking on the vertex, and then on another point close to the vertex that fits on the axes. \[ y=x^{2}-10 x+21 \] 1. State whether the parabola opens up or down? OUpO Down 2. State the vertex as an ordered pair: $\square$ 3. State the $y$-intercept as an ordered pair $\square$ 4. Write the Equation of the Axis of Symmetry $\square$ 5. Click on the vertex and then another point of the parabola that fits on the axes below. Clear All Draw: $\square$
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Solution

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Solution Steps

Step 1: Determine if the parabola opens upwards or downwards

The given equation is $y = x^2 - 10x + 21$. Since the coefficient of the $x^2$ term is positive (1), the parabola opens upwards.

Step 2: Find the vertex

To find the vertex, we can complete the square or use the formula for the x-coordinate of the vertex, $x = -\frac{b}{2a}$, where $a=1$ and $b=-10$. $x = -\frac{-10}{2(1)} = \frac{10}{2} = 5$

Now substitute $x=5$ back into the equation to find the y-coordinate of the vertex: $y = (5)^2 - 10(5) + 21 = 25 - 50 + 21 = -4$ So, the vertex is $(5, -4)$.

Step 3: Find the y-intercept

The y-intercept occurs when $x=0$. Substitute $x=0$ into the equation: $y = (0)^2 - 10(0) + 21 = 21$ So the y-intercept is $(0, 21)$.

Final Answer:

  1. Up
  2. (5, -4)
  3. (0, 21)
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