Question

a. draw the line of symmetry on the graph.
b. vertex :
c. x intercepts :
d. y intercept :

Explanation:

Since the graph is not provided, we'll assume a general quadratic - function approach. For a quadratic function \(y = ax^{2}+bx + c\), the line of symmetry is given by \(x=-\frac{b}{2a}\), the vertex is \((-\frac{b}{2a}, f(-\frac{b}{2a}))\), the \(x\) - intercepts are found by setting \(y = 0\) and solving \(ax^{2}+bx + c=0\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), and the \(y\) - intercept is found by setting \(x = 0\) (so \(y=c\)).

Step1: Recall line - of - symmetry formula

For a quadratic function \(y = ax^{2}+bx + c\), the line of symmetry is \(x =-\frac{b}{2a}\). But since no function is given, we can't draw it numerically. However, conceptually, for a parabola, it is a vertical line passing through the vertex.

Step2: Recall vertex formula

The \(x\) - coordinate of the vertex of a quadratic function \(y=ax^{2}+bx + c\) is \(x =-\frac{b}{2a}\). To find the \(y\) - coordinate, we substitute \(x =-\frac{b}{2a}\) into the function \(y = ax^{2}+bx + c\). So the vertex is \((-\frac{b}{2a},a(-\frac{b}{2a})^{2}+b(-\frac{b}{2a})+c)\).

Step3: Recall \(x\) - intercept formula

Set \(y = 0\) in \(y=ax^{2}+bx + c\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), the \(x\) - intercepts are \((\frac{-b + \sqrt{b^{2}-4ac}}{2a},0)\) and \((\frac{-b-\sqrt{b^{2}-4ac}}{2a},0)\).

Step4: Recall \(y\) - intercept formula

Set \(x = 0\) in \(y=ax^{2}+bx + c\). Then \(y=c\), so the \(y\) - intercept is \((0,c)\).

Since no specific function is given:

Answer:

a. Without a specific function or graph, cannot be drawn numerically.
b. \((-\frac{b}{2a},a(-\frac{b}{2a})^{2}+b(-\frac{b}{2a})+c)\)
c. \((\frac{-b + \sqrt{b^{2}-4ac}}{2a},0)\), \((\frac{-b-\sqrt{b^{2}-4ac}}{2a},0)\)
d. \((0,c)\)