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Math  /  Algebra

QuestionDetermine the quadratic function whose graph is given below.
The quadratic function which describes the given graph is f(x)=f(x)= □\square □\square. (Type an expression.)

Studdy Solution

STEP 1

1. The quadratic function is in the form f(x)=ax2+bx+c f(x) = ax^2 + bx + c .
2. The vertex form of a quadratic function is f(x)=a(x−h)2+k f(x) = a(x-h)^2 + k , where (h,k)(h, k) is the vertex.
3. The parabola opens upwards, indicating that the coefficient a a is positive.

STEP 2

1. Use the vertex form of the quadratic function.
2. Substitute the vertex into the vertex form.
3. Use another point to solve for the coefficient a a .
4. Write the quadratic function in standard form.

STEP 3

Start with the vertex form of a quadratic function:
f(x)=a(x−h)2+k f(x) = a(x-h)^2 + k
Given the vertex (h,k)=(2,−5)(h, k) = (2, -5), substitute these values into the vertex form:
f(x)=a(x−2)2−5 f(x) = a(x-2)^2 - 5

STEP 4

Use the point (0,1)(0, 1) that lies on the parabola to solve for a a . Substitute x=0 x = 0 and f(x)=1 f(x) = 1 into the equation:
1=a(0−2)2−5 1 = a(0-2)^2 - 5

STEP 5

Simplify and solve for a a :
1=a(4)−5 1 = a(4) - 5 1=4a−5 1 = 4a - 5 6=4a 6 = 4a a=32 a = \frac{3}{2}

STEP 6

Substitute a=32 a = \frac{3}{2} back into the vertex form to get the quadratic function:
f(x)=32(x−2)2−5 f(x) = \frac{3}{2}(x-2)^2 - 5

STEP 7

Convert the function from vertex form to standard form by expanding:
f(x)=32(x−2)2−5 f(x) = \frac{3}{2}(x-2)^2 - 5 f(x)=32(x2−4x+4)−5 f(x) = \frac{3}{2}(x^2 - 4x + 4) - 5 f(x)=32x2−6x+6−5 f(x) = \frac{3}{2}x^2 - 6x + 6 - 5 f(x)=32x2−6x+1 f(x) = \frac{3}{2}x^2 - 6x + 1
The quadratic function is:
f(x)=32x2−6x+1 f(x) = \frac{3}{2}x^2 - 6x + 1

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