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14. find the vertex of the quadratic function represented by the table.…

Question

  1. find the vertex of the quadratic function represented by the table. explain your reasoning.
x-5-2147
g(x)-68-234134
  1. write a quadratic function in standard form with axis of symmetry x = -5 and y - intercept 3.
  1. you kick a kickball. the path of the ball is represented by y = x(0.6 - 0.02x), where x is the horizontal distance (in feet) and y is the corresponding height (in feet). your second kick reaches a maximum height of 7 feet, 12 feet away from you. which kick travels farther before hitting the ground? which kick travels higher?

Explanation:

Step1: Recall vertex property of quadratic

The vertex of a quadratic function lies on the axis of symmetry. For a quadratic function, the function values are symmetric about the axis of symmetry.

Step2: Identify symmetric points

We see that \(g(1) = 4\) and \(g(7)=4\). The axis of symmetry is \(x=\frac{1 + 7}{2}\).

Step3: Calculate axis - of - symmetry

\[x=\frac{1+7}{2}=\frac{8}{2}=4\]
When \(x = 4\), \(g(4)=13\). So the vertex of the quadratic function is \((4,13)\) since the vertex lies on the axis of symmetry and we can read the \(y\) - value from the table at \(x = 4\).

for Question 15:

Step1: Recall standard form of quadratic

The standard form of a quadratic function is \(y=ax^{2}+bx + c\). The axis of symmetry is given by \(x =-\frac{b}{2a}\), and the \(y\) - intercept is \(c\).

Step2: Use given information

We know that \(x=-\frac{b}{2a}=-5\), so \(b = 10a\), and \(c = 3\). Let \(a = 1\) (we can choose any non - zero value for \(a\)). Then \(b = 10\).

Step3: Write the quadratic function

The quadratic function is \(y=x^{2}+10x + 3\)

for Question 16:

Step1: Rewrite first kick function

For the first kick, \(y=x(0.6 - 0.02x)=-0.02x^{2}+0.6x\). The \(x\) - intercepts are found by setting \(y = 0\). So \(x(0.6 - 0.02x)=0\), which gives \(x = 0\) or \(x=\frac{0.6}{0.02}=30\). The maximum of \(y=-0.02x^{2}+0.6x\) occurs at \(x=-\frac{0.6}{2\times(-0.02)}=\frac{- 0.6}{-0.04}=15\), and \(y=-0.02\times15^{2}+0.6\times15=-0.02\times225 + 9=-4.5 + 9 = 4.5\)

Step2: Analyze second kick

The second kick has a vertex at \((12,7)\). The \(x\) - intercepts of a quadratic function with vertex \((h,k)\) in vertex form \(y=a(x - h)^{2}+k\) (we know \(h = 12,k = 7\)). Since the parabola is symmetric about \(x = 12\), the \(x\) - intercepts are symmetric about \(x = 12\). The ball hits the ground when \(y = 0\). The distance from the origin to the non - zero \(x\) - intercept is greater than \(12\) (because of symmetry). Let's assume the quadratic function of the second kick is \(y=a(x - 12)^{2}+7\). When \(y = 0\), \(a(x - 12)^{2}=-7\). Since the parabola is symmetric, the non - zero \(x\) - intercept is at a distance greater than \(12\) from the origin.

Step3: Compare

The first kick travels \(30\) feet horizontally before hitting the ground and has a maximum height of \(4.5\) feet. The second kick has a maximum height of \(7\) feet and travels more than \(24\) feet horizontally (since symmetric about \(x = 12\)).
The first kick travels farther before hitting the ground (\(30>24\)) and the second kick travels higher (\(7>4.5\))

Answer:

\((4,13)\)