Let's take a look at the graph of the function

y = a x^{2} + c

when

a

and

c

have the same sign. There are two possible cases.

Case I : Case II : a > 0, c > 0 a < 0, c < 0

Let's analyze them one at a time!

Case I

When

a > 0

the parabola opens upwards. Therefore, the function reaches its minimum value at the vertex. Let's rewrite our function in the vertex form.

y = a x^{2} + c \Leftrightarrow y = a (x - 0)^{2} + c

The vertex of the parabola lies at

(0, c) .

The minimum value of the function is

c > 0 .

Since the function never reaches the value of

0,

it cannot have

x -

intercepts.

Case II

We will follow a similar reasoning to Case I. When

a < 0

the parabola opens downwards. Therefore, the function reaches its maximum value at the vertex. Let's use the vertex form of our function.

y = a (x - 0)^{2} + c

Again, the vertex of the parabola lies at

(0, c) .

The maximum value of the function is

c < 0 .

Since the function never reaches the value of

0,

it cannot have

x -

intercepts.

Conclusion

We have shown that when

a

and

c

have the same sign the function

y = a x^{2} + c

has no

x -

intercepts. Therefore, the given statement is always true.

Alternative Solution

Algebraic Approach

We will try to find the

x -

intercepts of the graph of

y = a x^{2} + c

when

a

and

c

have the same sign. This means finding the

x -

values corresponding to

y = 0 .

0 = a x^{2} + c

Let's transform this equation!

0 = a x^{2} + c

RearrangeEqn

Rearrange equation

a x^{2} + c = 0

SubEqn

$LHS - c = RHS - c$

a x^{2} = - c

DivEqn

$LHS / a = RHS / a$

x^{2} = \frac{- c}{a}

MoveNegNumToFrac

Put minus sign in front of fraction

x^{2} = - \frac{c}{a}

Since

a

and

c

have the same sign the quotient

\frac{c}{a}

is positive. Thus,

- \frac{c}{a}

is always negative. We also know that any real number, other than

0,

squared, is positive. Therefore, the equation

x^{2} = - \frac{c}{a}

has no real solutions and the function

y = a x^{2} + c

has no

x -

intercepts when

a

and

c

have the same sign. Therefore, the statement is always true.

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