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Question
question 6 (mandatory) (1 point)
what information can you gather immediately from a quadratic function written in standard form?
a) equation of the axis of symmetry
b) y-intercept
c) domain
d) vertex
Brief Explanations
- Recall the standard form of a quadratic function: The standard form of a quadratic function is \( y = ax^{2}+bx + c\), where \(a
eq0\).
- Option a: The formula for the axis of symmetry of a quadratic function \(y = ax^{2}+bx + c\) is \(x=-\frac{b}{2a}\). We can calculate it using the coefficients \(a\) and \(b\) from the standard form, but we don't get it "immediately" just by looking at the standard form (we need to do a calculation).
- Option b: To find the \(y\)-intercept, we set \(x = 0\). When \(x=0\), \(y=c\). So from the standard form \(y=ax^{2}+bx + c\), we can immediately see that the \(y\)-intercept is \(c\) (we just look at the constant term).
- Option c: The domain of any quadratic function (a polynomial function) is all real numbers (\((-\infty,\infty)\)). This is a property of quadratic functions in general, not something we gather specifically from the standard form (it's true for all quadratics regardless of the values of \(a\), \(b\), and \(c\)).
- Option d: The vertex form of a quadratic function is \(y=a(x - h)^{2}+k\), where \((h,k)\) is the vertex. To find the vertex from the standard form \(y=ax^{2}+bx + c\), we need to complete the square or use the formula \(h=-\frac{b}{2a}\) and then find \(k\) by substituting \(h\) into the function. We don't get the vertex immediately from the standard form.
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b) \( y \)-intercept