Question

find the vertex of the quadratic passing through the point f(3/2)=9, has an x-intercept of 3 and the axis of symmetry is x=2

Answer 1

The vertex of the quadratic is (2 , 12)

Step-by-step explanation:

The vertex form of the quadratic function is f(x) = a(x - h)² + k, where

• (h , k) are the coordinates of its vertex
• Its axis of symmetry is a vertical line at x = h
• its minimum value is f(x) = k at x = h

∵ The function of quadratic

∴ f(x) = a(x - h)² + k

∵ The axis of symmetry of the function is x = 2

∵ The axis of symmetry of a quadratic function is a vertical line at x = h

∴ h = 2

- Substitute h by 2 in f(x)

∴ f(x) = a(x - 2)² + k

∵ The quadratic passing through the point f(3/2) = 9

- That means the coordinates of the point are $(\frac{3}{2},9)$

∴ x = $\frac{3}{2}$  and y = 9

- Substitute x and y in the equation by the values above

∴ 9 = a( $\frac{3}{2}$ - 2)² + k

∴ 9 = a( $\frac{-1}{2}$ )² + k

∴ 9 = $\frac{1}{4}$ a + k

- Multiply each term by 4

∴ 36 = a + 4 k

∴ a + 4 k = 36 ⇒ (1)

∵ x intercept of f(x) = 3

- That means the graph of f(x) intersects x-axis at point (3 , 0)

∴ x = 3 and y = 0

- Substitute x and y in the equation by the values above

∴ 0 = a(3 - 2)² + k

∴ 0 = a(1)² + k

∴ 0 = a + k

∴ a + k = 0 ⇒ (2)

Now we have system of equation let us solve it to find k

∵ a + 4 k = 36 ⇒ (1)

∵ a + k = 0 ⇒ (2)

- Subtract equation (2) from equation (1) to eliminate a

∴ 3 k = 36

- Divide both sides by 3

∴ k = 12

∵ (h , k) are the coordinates of the vertex of the quadratic function

∴ (2 , 12) are the coordinates of the vertex of the quadratic function

The vertex of the quadratic is (2 , 12)

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