Standard Form of a Quadratic Find the vertex of a quadratic function f(x) = 3x^2 + 18x + 32. Find the vertex, axis of symmetry, and the graph of f(x) = 4x^2 – 8x + 19. Vertex, Axis of Symmetry, and Graph.
The Correct Answer and Explanation is:
Problem 1:
Find the vertex of the quadratic function
f(x)=3×2+18x+32f(x) = 3x^2 + 18x + 32f(x)=3×2+18x+32
To find the vertex of a quadratic function in standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c, we use the formula for the x-value of the vertex:x=−b2ax = \frac{-b}{2a}x=2a−b
Here,
a=3a = 3a=3,
b=18b = 18b=18x=−182⋅3=−186=−3x = \frac{-18}{2 \cdot 3} = \frac{-18}{6} = -3x=2⋅3−18=6−18=−3
Now substitute x=−3x = -3x=−3 into the original function to find the y-value:f(−3)=3(−3)2+18(−3)+32=3(9)−54+32=27−54+32=5f(-3) = 3(-3)^2 + 18(-3) + 32 = 3(9) – 54 + 32 = 27 – 54 + 32 = 5f(−3)=3(−3)2+18(−3)+32=3(9)−54+32=27−54+32=5
So the vertex is:(−3,5)\boxed{(-3, 5)}(−3,5)
Problem 2:
Find the vertex, axis of symmetry, and graph the function
f(x)=4×2−8x+19f(x) = 4x^2 – 8x + 19f(x)=4×2−8x+19
Again, use the vertex formula:x=−b2a=−(−8)2⋅4=88=1x = \frac{-b}{2a} = \frac{-(-8)}{2 \cdot 4} = \frac{8}{8} = 1x=2a−b=2⋅4−(−8)=88=1
Substitute x=1x = 1x=1 into the function to find the y-value:f(1)=4(1)2−8(1)+19=4−8+19=15f(1) = 4(1)^2 – 8(1) + 19 = 4 – 8 + 19 = 15f(1)=4(1)2−8(1)+19=4−8+19=15
So the vertex is:(1,15)\boxed{(1, 15)}(1,15)
The axis of symmetry is the vertical line that passes through the x-value of the vertex:x=1\boxed{x = 1}x=1
Summary and Explanation (300 words)
Quadratic functions in standard form are written as f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c. The shape of the graph is a parabola. If a>0a > 0a>0, the parabola opens upward. If a<0a < 0a<0, it opens downward. The vertex is the turning point of the parabola and is found using the formula x=−b2ax = \frac{-b}{2a}x=2a−b. Once the x-value is found, plug it back into the function to find the corresponding y-value. This gives the coordinates of the vertex.
In the first example, the function f(x)=3×2+18x+32f(x) = 3x^2 + 18x + 32f(x)=3×2+18x+32 has a vertex at (−3,5)(-3, 5)(−3,5). This point represents the minimum value of the function since the parabola opens upward. The second example, f(x)=4×2−8x+19f(x) = 4x^2 – 8x + 19f(x)=4×2−8x+19, has a vertex at (1,15)(1, 15)(1,15) and also opens upward because the coefficient of x2x^2×2 is positive. The axis of symmetry is simply a vertical line that passes through the x-value of the vertex. It divides the graph into two symmetrical halves.
To graph either function, plot the vertex, draw the axis of symmetry, and then choose x-values on either side of the vertex to plot additional points. The symmetry of the graph helps to complete the shape of the parabola.
These methods are useful in algebra and real-world applications where finding maximum or minimum values is important, such as in business or physics problems.
